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  • Physics Formulas

Free Fall Formula

Freefall as the term says, is a body falling freely because of the gravitational pull of our earth.

Imagine a body with velocity (v) is falling freely from a height (h) for time (t) seconds because of gravity (g).

Free Fall

Free Fall Formulas  are articulated as follows:

h = (1/2) gt 2

Freefall Related Solved Examples

Underneath are given questions on free fall which may be useful for you.

Problem 1:  Calculate  the  body  height  if  it  has  a  mass  of  2  kg  and  after  7 seconds  it  reaches  the  ground?

Given: Height h =? Time t = 7s We all are acquainted with the fact that free fall is independent of mass.

Hence, it is given as

h = 0.5 × 9.8 × (7) 2

h = 240.1 m

Problem  2: The cotton falls after 3 s and iron falls after 5 s. Which is moving with higher velocity? Answer:

The Velocity in free fall is autonomous of mass.

V (Velocity of iron) = gt = 9.8 m/s 2  × 5s = 49 m/s

V (Velocity of cotton) = gt = 9.8 m/s 2  × 3s = 29.4 m/s.

The Velocity of iron is more than cotton.

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Application: Falling Objects (1D)

Application: projectile motion (2d), forces & newton's laws of motion, potential energy, thermal energy.

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Free Fall (Physics): Definition, Formula, Problems & Solutions (w/ Examples)

​ Free fall ​ refers to situations in physics where the only force acting on an object is gravity.

The simplest examples occur when objects fall from a given height above the surface of the Earth straight downward – a one-dimensional problem. If the object is tossed upward or forcefully thrown straight downward, the example is still one-dimensional, but with a twist.

Projectile motion is a classic category of free-fall problems. In reality, of course, these events unfold in the three-dimensional world, but for introductory physics purposes, they are treated on paper (or on your screen) as two-dimensional: ​ x ​ for right and left (with right being positive), and ​ y ​ for up and down (with up being positive).

Free-fall examples therefore often have negative values for y-displacement.

It is perhaps counterintuitive that some free-fall problems qualify as such.

Keep in mind that the only criterion is that the only force acting on the object is gravity (usually Earth's gravity). Even if an object is launched into the sky with colossal initial force, at the moment the object is released and thereafter, the only force acting on it is gravity and it is now a projectile.

  • Often, high-school and many college physics problems neglect air resistance, though this always has at least a slight effect in reality; the exception is an event that unfolds in a vacuum. This is discussed in detail later.

The Unique Contribution of Gravity

A unique an interesting property of the acceleration due to gravity is that it is the same for all masses.

This was far from self-evident until the days of Galileo Galilei (1564-1642). That's because in reality gravity is not the only force acting as an object falls, and the effects of air resistance tend to cause lighter objects to accelerate more slowly – something we've all noticed when comparing the fall rate of a rock and a feather.

Galileo conducted ingenious experiments at the "leaning" Tower of Pisa, proving by dropping masses of different weights from the high top of the tower that gravitational acceleration is independent of mass.

Solving Free-Fall Problems

Usually, you are looking to determine initial velocity (v 0y ), final velocity (v y ) or how far something has fallen (y − y 0 ). Although Earth's gravitational acceleration is a constant 9.8 m/s 2 , elsewhere (such as on the moon) the constant acceleration experienced by an object in free fall has a different value.

For free fall in one dimension (for example, an apple falling straight down from a tree), use the kinematic equations in the ​ Kinematic Equations for Free-Falling Objects ​ section. For a projectile-motion problem in two dimensions, use the kinematic equations in the section ​ Projectile Motion and Coordinate Systems ​.

  • You can also use the conservation of energy principle, which states that ​ the loss of potential energy (PE) ​ during the fall ​ equals the gain in kinetic energy (KE): ​ –mg(y − y 0 ) = (1/2)mv y 2 . 

Kinematic Equations for Free-Falling Objects

All of the foregoing can be reduced for present purposes to the following three equations. These are tailored for free fall, so that the "y" subscripts can be omitted. Assume that acceleration, per physics convention, equals −g (with the positive direction therefore upward).

  • Note that v 0 and y 0  are initial values in any problem, not variables.

​ Example 1: ​ A strange birdlike animal is hovering in the air 10 m directly over your head, daring you to hit it with the rotten tomato you're holding. With what minimum initial velocity v 0 would you have to throw the tomato straight up in order to ensure that it reaches its squawking target?

What's happening physically is that the ball is coming to a stop owing to the force of gravity just as it reaches the required height, so here, v y = v = 0.

First, list your known quantities: ​ v = ​ 0​ , g = ​ –9.8 m/s2​ , y − y 0 = ​ 10 m

Thus you can use the third of the equations above to solve:

This is about 31 miles an hour.

Projectile Motion and Coordinate Systems

Projectile motion involves the motion of an object in (usually) two dimensions under the force of gravity. The behavior of the object in the x-direction and in the y-direction can be described separately in assembling the greater picture of the particle's motion. This means that "g" appears in most of the equations required to solve all projectile-motion problems, not merely those involving free fall.

The kinematic equations needed to solve basic projectile motion problems, which omit air resistance:

​ Example 2: ​ A daredevil decides to try to drive his "rocket car" across the gap between adjacent building rooftops. These are separated by 100 horizontal meters, and the roof of the "take-off" building is 30 m higher than the second (this almost 100 feet, or perhaps 8 to 10 "floors," i.e., levels).

Neglecting air resistance, how fast will he need to be going as he leaves the first rooftop to assure just reaching the second rooftop? Assume his vertical velocity is zero at the instant the car takes off.

Again, list your known quantities: (x – x 0 ) = 100m, (y – y 0 ) = –30m, v 0y = 0, g = –9.8 m/s 2 .

Here, you take advantage of the fact that horizontal motion and vertical motion can be assessed independently. How long the car will take to free-fall (for purposes of y-motion) 30 m? The answer is given by y – y 0 = v 0y t − (1/2)gt 2.

Filling in the known quantities and solving for t:

Now plug this value into x = x 0 + v 0x t :

​ v 0x = 40.4 m/s (about 90 miles per hour). ​

This is perhaps possible, depending on the size of the roof, but all in all not a good idea outside of action-hero movies.

Hitting it out of the Park... Far Out

Air resistance plays a major, under-appreciated role in everyday events even when free fall is only part of the physical story. In 2018, a professional baseball player named Giancarlo Stanton hit a pitched ball hard enough to blast it away from home plate at a record 121.7 miles per hour.

The equation for the maximum horizontal distance a launched projectile can attain, or ​ range equation ​ (see Resources), is:

Based on this, if Stanton had hit the ball at the theoretical ideal angle of 45 degrees (where sin 2θ is at its maximum value of 1), the ball would have traveled 978 feet! In reality, home runs almost never reach even 500 feet. Part if this is because a launch angle of 45 degrees for a batter is not ideal, as the pitch is coming in almost horizontally. But much of the difference is owed to the velocity-dampening effects of air resistance.

Air Resistance: Anything But "Negligible"

Free-fall physics problems aimed at less advanced students assume the absence of air resistance because this factor would introduce another force that can slow or decelerate objects and would need to be mathematically accounted for. This is a task best reserved for advanced courses, but it bears discussion here nonetheless.

In the real world, the Earth's atmosphere provides some resistance to an object in free fall. Particles in the air collide with the falling object, which results in transforming some of its kinetic energy into thermal energy. Since energy is conserved in general, this results in "less motion" or a more slowly increasing downward velocity.

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About the Author

Kevin Beck holds a bachelor's degree in physics with minors in math and chemistry from the University of Vermont. Formerly with ScienceBlogs.com and the editor of "Run Strong," he has written for Runner's World, Men's Fitness, Competitor, and a variety of other publications. More about Kevin and links to his professional work can be found at www.kemibe.com.

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Projectile Motion (Physics): Definition, Equations, Problems (w/ Examples)

Newton's laws of motion: what are they & why they matter, momentum (physics): definition, equation, units (w/ diagrams & examples), law of conservation of mass: definition, formula, history (w/ examples), potential energy: what is it & why it matters (w/ formula & examples), thermal energy: definition, equation, types (w/ diagram & examples).

  • 3.5 Free Fall
  • Introduction
  • 1.1 The Scope and Scale of Physics
  • 1.2 Units and Standards
  • 1.3 Unit Conversion
  • 1.4 Dimensional Analysis
  • 1.5 Estimates and Fermi Calculations
  • 1.6 Significant Figures
  • 1.7 Solving Problems in Physics
  • Key Equations
  • Conceptual Questions
  • Additional Problems
  • Challenge Problems
  • 2.1 Scalars and Vectors
  • 2.2 Coordinate Systems and Components of a Vector
  • 2.3 Algebra of Vectors
  • 2.4 Products of Vectors
  • 3.1 Position, Displacement, and Average Velocity
  • 3.2 Instantaneous Velocity and Speed
  • 3.3 Average and Instantaneous Acceleration
  • 3.4 Motion with Constant Acceleration
  • 3.6 Finding Velocity and Displacement from Acceleration
  • 4.1 Displacement and Velocity Vectors
  • 4.2 Acceleration Vector
  • 4.3 Projectile Motion
  • 4.4 Uniform Circular Motion
  • 4.5 Relative Motion in One and Two Dimensions
  • 5.2 Newton's First Law
  • 5.3 Newton's Second Law
  • 5.4 Mass and Weight
  • 5.5 Newton’s Third Law
  • 5.6 Common Forces
  • 5.7 Drawing Free-Body Diagrams
  • 6.1 Solving Problems with Newton’s Laws
  • 6.2 Friction
  • 6.3 Centripetal Force
  • 6.4 Drag Force and Terminal Speed
  • 7.2 Kinetic Energy
  • 7.3 Work-Energy Theorem
  • 8.1 Potential Energy of a System
  • 8.2 Conservative and Non-Conservative Forces
  • 8.3 Conservation of Energy
  • 8.4 Potential Energy Diagrams and Stability
  • 8.5 Sources of Energy
  • 9.1 Linear Momentum
  • 9.2 Impulse and Collisions
  • 9.3 Conservation of Linear Momentum
  • 9.4 Types of Collisions
  • 9.5 Collisions in Multiple Dimensions
  • 9.6 Center of Mass
  • 9.7 Rocket Propulsion
  • 10.1 Rotational Variables
  • 10.2 Rotation with Constant Angular Acceleration
  • 10.3 Relating Angular and Translational Quantities
  • 10.4 Moment of Inertia and Rotational Kinetic Energy
  • 10.5 Calculating Moments of Inertia
  • 10.6 Torque
  • 10.7 Newton’s Second Law for Rotation
  • 10.8 Work and Power for Rotational Motion
  • 11.1 Rolling Motion
  • 11.2 Angular Momentum
  • 11.3 Conservation of Angular Momentum
  • 11.4 Precession of a Gyroscope
  • 12.1 Conditions for Static Equilibrium
  • 12.2 Examples of Static Equilibrium
  • 12.3 Stress, Strain, and Elastic Modulus
  • 12.4 Elasticity and Plasticity
  • 13.1 Newton's Law of Universal Gravitation
  • 13.2 Gravitation Near Earth's Surface
  • 13.3 Gravitational Potential Energy and Total Energy
  • 13.4 Satellite Orbits and Energy
  • 13.5 Kepler's Laws of Planetary Motion
  • 13.6 Tidal Forces
  • 13.7 Einstein's Theory of Gravity
  • 14.1 Fluids, Density, and Pressure
  • 14.2 Measuring Pressure
  • 14.3 Pascal's Principle and Hydraulics
  • 14.4 Archimedes’ Principle and Buoyancy
  • 14.5 Fluid Dynamics
  • 14.6 Bernoulli’s Equation
  • 14.7 Viscosity and Turbulence
  • 15.1 Simple Harmonic Motion
  • 15.2 Energy in Simple Harmonic Motion
  • 15.3 Comparing Simple Harmonic Motion and Circular Motion
  • 15.4 Pendulums
  • 15.5 Damped Oscillations
  • 15.6 Forced Oscillations
  • 16.1 Traveling Waves
  • 16.2 Mathematics of Waves
  • 16.3 Wave Speed on a Stretched String
  • 16.4 Energy and Power of a Wave
  • 16.5 Interference of Waves
  • 16.6 Standing Waves and Resonance
  • 17.1 Sound Waves
  • 17.2 Speed of Sound
  • 17.3 Sound Intensity
  • 17.4 Normal Modes of a Standing Sound Wave
  • 17.5 Sources of Musical Sound
  • 17.7 The Doppler Effect
  • 17.8 Shock Waves
  • B | Conversion Factors
  • C | Fundamental Constants
  • D | Astronomical Data
  • E | Mathematical Formulas
  • F | Chemistry
  • G | The Greek Alphabet

Learning Objectives

By the end of this section, you will be able to:

  • Use the kinematic equations with the variables y and g to analyze free-fall motion.
  • Describe how the values of the position, velocity, and acceleration change during a free fall.
  • Solve for the position, velocity, and acceleration as functions of time when an object is in a free fall.

An interesting application of Equation 3.4 through Equation 3.14 is called free fall , which describes the motion of an object falling in a gravitational field, such as near the surface of Earth or other celestial objects of planetary size. Let’s assume the body is falling in a straight line perpendicular to the surface, so its motion is one-dimensional. For example, we can estimate the depth of a vertical mine shaft by dropping a rock into it and listening for the rock to hit the bottom. But “falling,” in the context of free fall, does not necessarily imply the body is moving from a greater height to a lesser height. If a ball is thrown upward, the equations of free fall apply equally to its ascent as well as its descent.

The most remarkable and unexpected fact about falling objects is that if air resistance and friction are negligible, then in a given location all objects fall toward the center of Earth with the same constant acceleration , independent of their mass . This experimentally determined fact is unexpected because we are so accustomed to the effects of air resistance and friction that we expect light objects to fall slower than heavy ones. Until Galileo Galilei (1564–1642) proved otherwise, people believed that a heavier object has a greater acceleration in a free fall. We now know this is not the case. In the absence of air resistance, heavy objects arrive at the ground at the same time as lighter objects when dropped from the same height Figure 3.26 .

In the real world, air resistance can cause a lighter object to fall slower than a heavier object of the same size. A tennis ball reaches the ground after a baseball dropped at the same time. (It might be difficult to observe the difference if the height is not large.) Air resistance opposes the motion of an object through the air, and friction between objects—such as between clothes and a laundry chute or between a stone and a pool into which it is dropped—also opposes motion between them.

For the ideal situations of these first few chapters, an object falling without air resistance or friction is defined to be in free fall . The force of gravity causes objects to fall toward the center of Earth. The acceleration of free-falling objects is therefore called acceleration due to gravity . Acceleration due to gravity is constant, which means we can apply the kinematic equations to any falling object where air resistance and friction are negligible. This opens to us a broad class of interesting situations.

Acceleration due to gravity is so important that its magnitude is given its own symbol, g . It is constant at any given location on Earth and has the average value

Although g varies from 9.78 m/s 2 to 9.83 m/s 2 , depending on latitude, altitude, underlying geological formations, and local topography, let’s use an average value of 9.8 m/s 2 rounded to two significant figures in this text unless specified otherwise. Neglecting these effects on the value of g as a result of position on Earth’s surface, as well as effects resulting from Earth’s rotation, we take the direction of acceleration due to gravity to be downward (toward the center of Earth). In fact, its direction defines what we call vertical. Note that whether acceleration a in the kinematic equations has the value + g or − g depends on how we define our coordinate system. If we define the upward direction as positive, then a = − g = −9.8 m/s 2 , a = − g = −9.8 m/s 2 , and if we define the downward direction as positive, then a = g = 9.8 m/s 2 a = g = 9.8 m/s 2 .

One-Dimensional Motion Involving Gravity

The best way to see the basic features of motion involving gravity is to start with the simplest situations and then progress toward more complex ones. So, we start by considering straight up-and-down motion with no air resistance or friction. These assumptions mean the velocity (if there is any) is vertical. If an object is dropped, we know the initial velocity is zero when in free fall. When the object has left contact with whatever held or threw it, the object is in free fall. When the object is thrown, it has the same initial speed in free fall as it did before it was released. When the object comes in contact with the ground or any other object, it is no longer in free fall and its acceleration of g is no longer valid. Under these circumstances, the motion is one-dimensional and has constant acceleration of magnitude g . We represent vertical displacement with the symbol y .

Kinematic Equations for Objects in Free Fall

We assume here that acceleration equals − g (with the positive direction upward).

Problem-Solving Strategy

  • Decide on the sign of the acceleration of gravity. In Equation 3.15 through Equation 3.17 , acceleration g is negative, which says the positive direction is upward and the negative direction is downward. In some problems, it may be useful to have acceleration g as positive, indicating the positive direction is downward.
  • Draw a sketch of the problem. This helps visualize the physics involved.
  • Record the knowns and unknowns from the problem description. This helps devise a strategy for selecting the appropriate equations to solve the problem.
  • Decide which of Equation 3.15 through Equation 3.17 are to be used to solve for the unknowns.

Example 3.14

Free fall of a ball.

  • Substitute the given values into the equation: y = y 0 + v 0 t − 1 2 g t 2 − 98.0 m = 0 − ( 4.9 m/s ) t − 1 2 ( 9.8 m/s 2 ) t 2 . y = y 0 + v 0 t − 1 2 g t 2 − 98.0 m = 0 − ( 4.9 m/s ) t − 1 2 ( 9.8 m/s 2 ) t 2 . This simplifies to t 2 + t − 20 = 0 . t 2 + t − 20 = 0 . This is a quadratic equation with roots t = −5.0 s and t = 4.0 s t = −5.0 s and t = 4.0 s . The positive root is the one we are interested in, since time t = 0 t = 0 is the time when the ball is released at the top of the building. (The time t = −5.0 s t = −5.0 s represents the fact that a ball thrown upward from the ground would have been in the air for 5.0 s when it passed by the top of the building moving downward at 4.9 m/s.)
  • Using Equation 3.15 , we have v = v 0 − g t = −4.9 m/s − ( 9.8 m/s 2 ) ( 4.0 s ) = −44.1 m/s . v = v 0 − g t = −4.9 m/s − ( 9.8 m/s 2 ) ( 4.0 s ) = −44.1 m/s .

Significance

Example 3.15, vertical motion of a baseball.

  • Equation 3.16 gives y = y 0 + v 0 t − 1 2 g t 2 y = y 0 + v 0 t − 1 2 g t 2 0 = 0 + v 0 ( 5.0 s ) − 1 2 ( 9.8 m / s 2 ) ( 5.0 s ) 2 , 0 = 0 + v 0 ( 5.0 s ) − 1 2 ( 9.8 m / s 2 ) ( 5.0 s ) 2 , which gives v 0 = 24.5 m/s v 0 = 24.5 m/s .
  • At the maximum height, v = 0 v = 0 . With v 0 = 24.5 m/s v 0 = 24.5 m/s , Equation 3.17 gives v 2 = v 0 2 − 2 g ( y − y 0 ) v 2 = v 0 2 − 2 g ( y − y 0 ) 0 = ( 24.5 m/s ) 2 − 2 ( 9.8 m/s 2 ) ( y − 0 ) 0 = ( 24.5 m/s ) 2 − 2 ( 9.8 m/s 2 ) ( y − 0 ) or y = 30.6 m . y = 30.6 m .
  • To find the time when v = 0 v = 0 , we use Equation 3.15 : v = v 0 − g t v = v 0 − g t 0 = 24.5 m/s − ( 9.8 m/s 2 ) t . 0 = 24.5 m/s − ( 9.8 m/s 2 ) t . This gives t = 2.5 s t = 2.5 s . Since the ball rises for 2.5 s, the time to fall is 2.5 s.
  • The acceleration is 9.8 m/s 2 everywhere, even when the velocity is zero at the top of the path. Although the velocity is zero at the top, it is changing at the rate of 9.8 m/s 2 downward.
  • The velocity at t = 5.0 s t = 5.0 s can be determined with Equation 3.15 : v = v 0 − g t = 24.5 m/s − 9.8 m/s 2 ( 5.0 s ) = −24.5 m/s . v = v 0 − g t = 24.5 m/s − 9.8 m/s 2 ( 5.0 s ) = −24.5 m/s .

Check Your Understanding 3.7

A chunk of ice breaks off a glacier and falls 30.0 m before it hits the water. Assuming it falls freely (there is no air resistance), how long does it take to hit the water? Which quantity increases faster, the speed of the ice chunk or its distance traveled?

Example 3.16

Rocket booster.

  • From Equation 3.17 , v 2 = v 0 2 − 2 g ( y − y 0 ) v 2 = v 0 2 − 2 g ( y − y 0 ) . With v = 0 and y 0 = 0 v = 0 and y 0 = 0 , we can solve for y : y = v 0 2 2 g = ( 2.0 × 10 2 m / s ) 2 2 ( 9.8 m / s 2 ) = 2040.8 m . y = v 0 2 2 g = ( 2.0 × 10 2 m / s ) 2 2 ( 9.8 m / s 2 ) = 2040.8 m . This solution gives the maximum height of the booster in our coordinate system, which has its origin at the point of release, so the maximum height of the booster is roughly 7.0 km.
  • An altitude of 6.0 km corresponds to y = 1.0 × 10 3 m y = 1.0 × 10 3 m in the coordinate system we are using. The other initial conditions are y 0 = 0 , and v 0 = 200.0 m/s y 0 = 0 , and v 0 = 200.0 m/s . We have, from Equation 3.17 , v 2 = ( 200.0 m / s ) 2 − 2 ( 9.8 m / s 2 ) ( 1.0 × 10 3 m ) ⇒ v = ± 142.8 m / s . v 2 = ( 200.0 m / s ) 2 − 2 ( 9.8 m / s 2 ) ( 1.0 × 10 3 m ) ⇒ v = ± 142.8 m / s .

Interactive

Engage the Phet simulation below to learn about graphing polynomials. The shape of the curve changes as the constants are adjusted. View the curves for the individual terms (for example, y = bx ) to see how they add to generate the polynomial curve.

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free fall problem with solution

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free fall problem with solution

Like any moving object, the motion of an object in free fall can be described by four kinematic equations. The kinematic equations that describe any object's motion are:

The symbols in the above equation have a specific meaning: the symbol d stands for the displacement ; the symbol t stands for the time ; the symbol a stands for the acceleration of the object; the symbol v i stands for the initial velocity value; and the symbol v f stands for the final velocity .

Applying Free Fall Concepts to Problem-Solving

There are a few conceptual characteristics of free fall motion that will be of value when using the equations to analyze free fall motion. These concepts are described as follows:

  • An object in free fall experiences an acceleration of -9.8 m/s/s. (The - sign indicates a downward acceleration.) Whether explicitly stated or not, the value of the acceleration in the kinematic equations is -9.8 m/s/s for any freely falling object.
  • If an object is merely dropped (as opposed to being thrown) from an elevated height, then the initial velocity of the object is 0 m/s.
  • If an object is projected upwards in a perfectly vertical direction, then it will slow down as it rises upward. The instant at which it reaches the peak of its trajectory, its velocity is 0 m/s. This value can be used as one of the motion parameters in the kinematic equations; for example, the final velocity ( v f ) after traveling to the peak would be assigned a value of 0 m/s.
  • If an object is projected upwards in a perfectly vertical direction, then the velocity at which it is projected is equal in magnitude and opposite in sign to the velocity that it has when it returns to the same height. That is, a ball projected vertically with an upward velocity of +30 m/s will have a downward velocity of -30 m/s when it returns to the same height.

These four principles and the four kinematic equations can be combined to solve problems involving the motion of free falling objects. The two examples below illustrate application of free fall principles to kinematic problem-solving. In each example, the problem solving strategy that was introduced earlier in this lesson will be utilized.  

Example Problem A

Luke Autbeloe drops a pile of roof shingles from the top of a roof located 8.52 meters above the ground. Determine the time required for the shingles to reach the ground.

The solution to this problem begins by the construction of an informative diagram of the physical situation. This is shown below. The second step involves the identification and listing of known information in variable form. You might note that in the statement of the problem, there is only one piece of numerical information explicitly stated: 8.52 meters. The displacement ( d ) of the shingles is -8.52 m. (The - sign indicates that the displacement is downward). The remaining information must be extracted from the problem statement based upon your understanding of the above principles . For example, the v i value can be inferred to be 0 m/s since the shingles are dropped (released from rest; see note above ). And the acceleration ( a ) of the shingles can be inferred to be -9.8 m/s 2 since the shingles are free-falling ( see note above ). (Always pay careful attention to the + and - signs for the given quantities.) The next step of the solution involves the listing of the unknown (or desired) information in variable form. In this case, the problem requests information about the time of fall. So t is the unknown quantity. The results of the first three steps are shown in the table below.

The next step involves identifying a kinematic equation that allows you to determine the unknown quantity. There are four kinematic equations to choose from. In general, you will always choose the equation that contains the three known and the one unknown variable. In this specific case, the three known variables and the one unknown variable are d , v i , a , and t . Thus, you will look for an equation that has these four variables listed in it. An inspection of the four equations above reveals that the equation on the top left contains all four variables.

Once the equation is identified and written down, the next step involves substituting known values into the equation and using proper algebraic steps to solve for the unknown information. This step is shown below.

-8.52 m = (0 m/s) • (t) + ½ • (-9.8 m/s 2 ) • (t) 2

-8.52 m = (0 m) *(t) + (-4.9 m/s 2 ) • (t) 2

-8.52 m = (-4.9 m/s 2 ) • (t) 2

(-8.52 m)/(-4.9 m/s 2 ) = t 2

1.739 s 2 = t 2

The solution above reveals that the shingles will fall for a time of 1.32 seconds before hitting the ground. (Note that this value is rounded to the third digit.)

The last step of the problem-solving strategy involves checking the answer to assure that it is both reasonable and accurate. The value seems reasonable enough. The shingles are falling a distance of approximately 10 yards (1 meter is pretty close to 1 yard); it seems that an answer between 1 and 2 seconds would be highly reasonable. The calculated time easily falls within this range of reasonability. Checking for accuracy involves substituting the calculated value back into the equation for time and insuring that the left side of the equation is equal to the right side of the equation. Indeed it is!  

Example Problem B

Rex Things throws his mother's crystal vase vertically upwards with an initial velocity of 26.2 m/s. Determine the height to which the vase will rise above its initial height.

Once more, the solution to this problem begins by the construction of an informative diagram of the physical situation. This is shown below. The second step involves the identification and listing of known information in variable form. You might note that in the statement of the problem, there is only one piece of numerical information explicitly stated: 26.2 m/s. The initial velocity ( v i ) of the vase is +26.2 m/s. (The + sign indicates that the initial velocity is an upwards velocity). The remaining information must be extracted from the problem statement based upon your understanding of the above principles . Note that the v f value can be inferred to be 0 m/s since the final state of the vase is the peak of its trajectory ( see note above ). The acceleration ( a ) of the vase is -9.8 m/s 2 ( see note above ). The next step involves the listing of the unknown (or desired) information in variable form. In this case, the problem requests information about the displacement of the vase (the height to which it rises above its starting height). So d is the unknown information. The results of the first three steps are shown in the table below.

The next step involves identifying a kinematic equation that would allow you to determine the unknown quantity. There are four kinematic equations to choose from. Again, you will always search for an equation that contains the three known variables and the one unknown variable. In this specific case, the three known variables and the one unknown variable are v i , v f , a , and d . An inspection of the four equations above reveals that the equation on the top right contains all four variables.

v f 2 = v i 2 + 2 • a • d

(0 m/s) 2 = (26.2 m/s) 2 + 2 •(-9.8m/s 2 ) •d

0 m 2 /s 2 = 686.44 m 2 /s 2 + (-19.6 m/s 2 ) •d

(-19.6 m/s 2 ) • d = 0 m 2 /s 2 -686.44 m 2 /s 2

(-19.6 m/s 2 ) • d = -686.44 m 2 /s 2

d = (-686.44 m 2 /s 2 )/ (-19.6 m/s 2 )

The solution above reveals that the vase will travel upwards for a displacement of 35.0 meters before reaching its peak. (Note that this value is rounded to the third digit.)

The last step of the problem-solving strategy involves checking the answer to assure that it is both reasonable and accurate. The value seems reasonable enough. The vase is thrown with a speed of approximately 50 mi/hr (merely approximate 1 m/s to be equivalent to 2 mi/hr). Such a throw will never make it further than one football field in height (approximately 100 m), yet will surely make it past the 10-yard line (approximately 10 meters). The calculated answer certainly falls within this range of reasonability. Checking for accuracy involves substituting the calculated value back into the equation for displacement and insuring that the left side of the equation is equal to the right side of the equation. Indeed it is!

Kinematic equations provide a useful means of determining the value of an unknown motion parameter if three motion parameters are known. In the case of a free-fall motion, the acceleration is often known. And in many cases, another motion parameter can be inferred through a solid knowledge of some basic kinematic principles . The next part of Lesson 6 provides a wealth of practice problems with answers and solutions.  

Physics Problems with Solutions

Free fall motion: tutorials with examples and solutions.

Problems on free fall motion are presented along with detailed solutions.

From rest, a car accelerated at 8 m/s 2 for 10 seconds. a) What is the position of the car at the end of the 10 seconds? b) What is the velocity of the car at the end of the 10 seconds? Solution to Problem 1

With an initial velocity of 20 km/h, a car accelerated at 8 m/s 2 for 10 seconds. a) What is the position of the car at the end of the 10 seconds? b) What is the velocity of the car at the end of the 10 seconds? Solution to Problem 2

A car accelerates uniformly from 0 to 72 km/h in 11.5 seconds. a) What is the acceleration of the car in m/s 2 ? b) What is the position of the car by the time it reaches the velocity of 72 km/h? Solution to Problem 3

An object is thrown straight down from the top of a building at a speed of 20 m/s. It hits the ground with a speed of 40 m/s. a) How high is the building? b) How long was the object in the air? Solution to Problem 4

A train brakes from 40 m/s to a stop over a distance of 100 m. a) What is the acceleration of the train? b) How much time does it take the train to stop? Solution to Problem 5

A boy on a bicycle increases his velocity from 5 m/s to 20 m/s in 10 seconds. a) What is the acceleration of the bicycle? b) What distance was covered by the bicycle during the 10 seconds? Solution to Problem 6

a) How long does it take an airplane to take off if it needs to reach a speed on the ground of 350 km/h over a distance of 600 meters (assume the plane starts from rest)? b) What is the acceleration of the airplane over the 600 meters? Solution to Problem 7

Starting from a distance of 20 meters to the left of the origin and at a velocity of 10 m/s, an object accelerates to the right of the origin for 5 seconds at 4 m/s 2 . What is the position of the object at the end of the 5 seconds of acceleration? Solution to Problem 8

What is the smallest distance, in meters, needed for an airplane touching the runway with a velocity of 360 km/h and an acceleration of -10 m/s 2 to come to rest? Solution to Problem 9

Problem 10:

To approximate the height of a water well, Martha and John drop a heavy rock into the well. 8 seconds after the rock is dropped, they hear a splash caused by the impact of the rock on the water. What is the height of the well. (Speed of sound in air is 340 m/s). Solution to Problem 10

Problem 11:

A rock is thrown straight up and reaches a height of 10 m. a) How long was the rock in the air? b) What is the initial velocity of the rock? Solution to Problem 11

Problem 12:

A car accelerates from rest at 1.0 m/s 2 for 20.0 seconds along a straight road . It then moves at a constant speed for half an hour. It then decelerates uniformly to a stop in 30.0 s. Find the total distance covered by the car. Solution to Problem 12

More References and links

  • Velocity and Speed: Tutorials with Examples
  • Velocity and Speed: Problems with Solutions
  • Acceleration: Tutorials with Examples
  • Uniform Acceleration Motion: Problems with Solutions
  • Uniform Acceleration Motion: Equations with Explanations

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Learning Objectives

  • Use the kinematic equations with the variables y and g to analyze free-fall motion.
  • Describe how the values of the position, velocity, and acceleration change during a free fall.
  • Solve for the position, velocity, and acceleration as functions of time when an object is in a free fall.
  • Use one-dimensional motion in perpendicular directions to analyze projectile motion.
  • Calculate the range, time of flight, and maximum height of a projectile that is launched and impacts a flat, horizontal surface.
  • Find the time of flight and impact velocity of a projectile that lands at a different height from that of launch.
  • Calculate the trajectory of a projectile.

An interesting application of Equation 3.3.2 through Equation 3.5.22 is called free fall , which describes the motion of an object falling in a gravitational field, such as near the surface of Earth or other celestial objects of planetary size. Let’s assume the body is falling in a straight line perpendicular to the surface, so its motion is one-dimensional. For example, we can estimate the depth of a vertical mine shaft by dropping a rock into it and listening for the rock to hit the bottom. But “falling,” in the context of free fall, does not necessarily imply the body is moving from a greater height to a lesser height. If a ball is thrown upward, the equations of free fall apply equally to its ascent as well as its descent.

The most remarkable and unexpected fact about falling objects is that if air resistance and friction are negligible, then in a given location all objects fall toward the center of Earth with the same constant acceleration, independent of their mass . This experimentally determined fact is unexpected because we are so accustomed to the effects of air resistance and friction that we expect light objects to fall slower than heavy ones. Until Galileo Galilei (1564–1642) proved otherwise, people believed that a heavier object has a greater acceleration in a free fall. We now know this is not the case. In the absence of air resistance, heavy objects arrive at the ground at the same time as lighter objects when dropped from the same height Figure \(\PageIndex{1}\).

Left figure shows a hammer and a feather falling down in air. Hammer is below the feather. Middle figure shows a hammer and a feather falling down in vacuum. Hammer and feather are at the same level. Right figure shows astronaut on the surface of the moon with hammer and a feather lying on the ground.

In the real world, air resistance can cause a lighter object to fall slower than a heavier object of the same size. A tennis ball reaches the ground after a baseball dropped at the same time. (It might be difficult to observe the difference if the height is not large.) Air resistance opposes the motion of an object through the air, and friction between objects—such as between clothes and a laundry chute or between a stone and a pool into which it is dropped—also opposes motion between them.

For the ideal situations of these first few chapters, an object falling without air resistance or friction is defined to be in free fall . The force of gravity causes objects to fall toward the center of Earth. The acceleration of free-falling objects is therefore called acceleration due to gravity . Acceleration due to gravity is constant, which means we can apply the kinematic equations to any falling object where air resistance and friction are negligible. This opens to us a broad class of interesting situations.

Acceleration due to gravity is so important that its magnitude is given its own symbol, g. It is constant at any given location on Earth and has the average value

\[g = 9.81\; m/s^{2}\; (or\; 32.2\; ft/s^{2}) \ldotp\]

Although g varies from 9.78 m/s 2 to 9.83 m/s 2 , depending on latitude, altitude, underlying geological formations, and local topography, let’s use an average value of 9.8 m/s 2 rounded to two significant figures in this text unless specified otherwise. Neglecting these effects on the value of g as a result of position on Earth’s surface, as well as effects resulting from Earth’s rotation, we take the direction of acceleration due to gravity to be downward (toward the center of Earth). In fact, its direction defines what we call vertical. Note that whether acceleration a in the kinematic equations has the value +g or -g depends on how we define our coordinate system. If we define the upward direction as positive, then a = -g = -9.8 m/s 2 , and if we define the downward direction as positive, then a = g = 9.8 m/s 2 .

One-Dimensional Motion Involving Gravity

The best way to see the basic features of motion involving gravity is to start with the simplest situations and then progress toward more complex ones. So, we start by considering straight up-and-down motion with no air resistance or friction. These assumptions mean the velocity (if there is any) is vertical. If an object is dropped, we know the initial velocity is zero when in free fall. When the object has left contact with whatever held or threw it, the object is in free fall. When the object is thrown, it has the same initial speed in free fall as it did before it was released. When the object comes in contact with the ground or any other object, it is no longer in free fall and its acceleration of g is no longer valid. Under these circumstances, the motion is one-dimensional and has constant acceleration of magnitude g. We represent vertical displacement with the symbol y.

Kinematic Equations for Objects in Free Fall

We assume here that acceleration equals -g (with the positive direction upward).

\[v =v _{0} - gt \label{3.15}\]

\[y = y_{0} + v_{0} t - \frac{1}{2} gt^{2} \label{3.16}\]

\[v^{2} = v_{0}^{2} - 2 g(y - y_{0}) \label{3.17}\]

Problem-Solving Strategy: Free Fall

  • Decide on the sign of the acceleration of gravity. In Equation \ref{3.15} through Equation \ref{3.17}, acceleration g is negative, which says the positive direction is upward and the negative direction is downward. In some problems, it may be useful to have acceleration g as positive, indicating the positive direction is downward.
  • Draw a sketch of the problem. This helps visualize the physics involved.
  • Record the knowns and unknowns from the problem description. This helps devise a strategy for selecting the appropriate equations to solve the problem.
  • Decide which of Equation \ref{3.15} through Equation \ref{3.17} are to be used to solve for the unknowns.

Example \(\PageIndex{1}\): Free Fall of a Ball

Figure \(\PageIndex{2}\) shows the positions of a ball, at 1-s intervals, with an initial velocity of 4.9 m/s downward, that is thrown from the top of a 98-m-high building. (a) How much time elapses before the ball reaches the ground? (b) What is the velocity when it arrives at the ground?

Figure shows the ball thrown downward from a tall building at a speed of - 4.9 meters per second. After one second, ball is lower by 9.8 meters and has a speed of -14.7 meters per second. After two seconds, ball is lower by 29.4 meters and has a speed of -24.5 meters per second. After three seconds, ball is lower by 58.8 meters and has a speed of -34.5 meters per second. After four seconds, ball is lower by 98.0 meters and has a speed of -44.1 meters per second.

Choose the origin at the top of the building with the positive direction upward and the negative direction downward. To find the time when the position is -98 m, we use Equation \ref{3.16}, with y 0 = 0, v 0 = -4.9 m/s, and g = 9.8 m/s 2 .

  • Substitute the given values into the equation: \(y = y_{0} + v_{0} t - \frac{1}{2} gt^{2}\) \(-98.0\; m = 0 - (4.9\; m/s)t - \frac{1}{2} (9.8\; m/s^{2}) t^{2} \ldotp\) This simplifies to \(t^{2} + t - 20 = 0 \ldotp\) This is a quadratic equation with roots t = -5.0 s and t = 4.0 s. The positive root is the one we are interested in, since time t = 0 is the time when the ball is released at the top of the building. (The time t = -5.0 s represents the fact that a ball thrown upward from the ground would have been in the air for 5.0 s when it passed by the top of the building moving downward at 4.9 m/s.)
  • Using Equation \ref{3.15}, we have \(v =v _{0} - gt = -4.9\; m/s - (9.8\; m/s^{2})(4.0\; s) = -44.1\; m/s \ldotp\)

Significance

For situations when two roots are obtained from a quadratic equation in the time variable, we must look at the physical significance of both roots to determine which is correct. Since t = 0 corresponds to the time when the ball was released, the negative root would correspond to a time before the ball was released, which is not physically meaningful. When the ball hits the ground, its velocity is not immediately zero, but as soon as the ball interacts with the ground, its acceleration is not g and it accelerates with a different value over a short time to zero velocity. This problem shows how important it is to establish the correct coordinate system and to keep the signs of g in the kinematic equations consistent.

Example \(\PageIndex{2}\): Vertical Motion of a Baseball

A batter hits a baseball straight upward at home plate and the ball is caught 5.0 s after it is struck Figure \(\PageIndex{3}\). (a) What is the initial velocity of the ball? (b) What is the maximum height the ball reaches? (c) How long does it take to reach the maximum height? (d) What is the acceleration at the top of its path? (e) What is the velocity of the ball when it is caught? Assume the ball is hit and caught at the same location.

Left picture shows a baseball player hitting the ball at time equal zero seconds. Right picture shows a baseball player catching the ball at time equal five seconds.

Choose a coordinate system with a positive y-axis that is straight up and with an origin that is at the spot where the ball is hit and caught.

  • Equation \ref{3.16} gives \(y = y_{0} + v_{0} t - \frac{1}{2} gt^{2}\) \(0 = 0 + v_{0} (5.0\; s)- \frac{1}{2} (9.8\; m/s^{2}) (5.0\; s)^{2} \ldotp\) which gives v 0 = 24.5 m/sec.
  • At the maximum height, v = 0. With v 0 = 24.5 m/s, Equation \ref{3.17} gives \(v^{2} = v_{0}^{2} - 2 g(y - y_{0})\) \(0 = (24.5\; m/s^{2}) - 2 (9.8\; m/s^{2})(y - 0)\) or \(y = 30.6\; m \ldotp\)
  • To find the time when v = 0 , we use Equation \ref{3.15}: \(v = v_{0} - gt\) \(0 = 24..5\; m/s - (9.8\; m/s^{2})t \ldotp\) This gives t = 2.5 s. Since the ball rises for 2.5 s, the time to fall is 2.5 s.
  • The acceleration is 9.8 m/s 2 everywhere, even when the velocity is zero at the top of the path. Although the velocity is zero at the top, it is changing at the rate of 9.8 m/s 2 downward.
  • The velocity at t = 5.0 s can be determined with Equation \ref{3.15}: \(\begin{split} v & = v_{0} - gt \\ & = 24.5\; m/s - 9.8\; m/s^{2} (5.0\; s) \\ & = -24.5\; m/s \ldotp \end{split}\)

The ball returns with the speed it had when it left. This is a general property of free fall for any initial velocity. We used a single equation to go from throw to catch, and did not have to break the motion into two segments, upward and downward. We are used to thinking of the effect of gravity is to create free fall downward toward Earth. It is important to understand, as illustrated in this example, that objects moving upward away from Earth are also in a state of free fall.

Example \(\PageIndex{3}\): Rocket Booster

A small rocket with a booster blasts off and heads straight upward. When at a height of 5.0 km and velocity of 200.0 m/s, it releases its booster. (a) What is the maximum height the booster attains? (b) What is the velocity of the booster at a height of 6.0 km? Neglect air resistance.

Figure shows a rocket releasing a booster.

We need to select the coordinate system for the acceleration of gravity, which we take as negative downward. We are given the initial velocity of the booster and its height. We consider the point of release as the origin. We know the velocity is zero at the maximum position within the acceleration interval; thus, the velocity of the booster is zero at its maximum height, so we can use this information as well. From these observations, we use Equation \ref{3.17}, which gives us the maximum height of the booster. We also use Equation \ref{3.17} to give the velocity at 6.0 km. The initial velocity of the booster is 200.0 m/s.

  • From Equation \ref{3.17}, \(v^{2} = v_{0}^{2} - 2 g(y - y_{0})\). With v = 0 and y 0 = 0, we can solve for y: \(y = \frac{v_{0}^{2}}{-2g} = \frac{(2.0 \times 10^{2}\; m/s)^{2}}{-2(9.8\; m/s^{2})} = 2040.8\; m \ldotp\) This solution gives the maximum height of the booster in our coordinate system, which has its origin at the point of release, so the maximum height of the booster is roughly 7.0 km.
  • An altitude of 6.0 km corresponds to y = 1.0 x 10 3 m in the coordinate system we are using. The other initial conditions are y 0 = 0, and v 0 = 200.0 m/s. We have, from Equation \ref{3.17}, \(v^{2} = (200.0\; m/s)^{2} - 2(9.8\; m/s^{2})(1.0 \times 10^{3}\; m) \Rightarrow v = \pm 142.8\; m/s \ldotp\)

We have both a positive and negative solution in (b). Since our coordinate system has the positive direction upward, the +142.8 m/s corresponds to a positive upward velocity at 6000 m during the upward leg of the trajectory of the booster. The value v = -142.8 m/s corresponds to the velocity at 6000 m on the downward leg. This example is also important in that an object is given an initial velocity at the origin of our coordinate system, but the origin is at an altitude above the surface of Earth, which must be taken into account when forming the solution.

Two-Dimensional Motion Involving Gravity-Projectile Motion

Projectile motion is the motion of an object thrown or projected into the air, subject only to acceleration as a result of gravity. The applications of projectile motion in physics and engineering are numerous. Some examples include meteors as they enter Earth’s atmosphere, fireworks, and the motion of any ball in sports. Such objects are called projectiles and their path is called a trajectory . The motion of falling objects as discussed in Motion Along a Straight Line is a simple one-dimensional type of projectile motion in which there is no horizontal movement. In this section, we consider two-dimensional projectile motion, and our treatment neglects the effects of air resistance.

The most important fact to remember here is that motions along perpendicular axes are independent and thus can be analyzed separately. We discussed this fact in Displacement and Velocity Vectors , where we saw that vertical and horizontal motions are independent. The key to analyzing two-dimensional projectile motion is to break it into two motions: one along the horizontal axis and the other along the vertical. (This choice of axes is the most sensible because acceleration resulting from gravity is vertical; thus, there is no acceleration along the horizontal axis when air resistance is negligible.) As is customary, we call the horizontal axis the x-axis and the vertical axis the y-axis. It is not required that we use this choice of axes; it is simply convenient in the case of gravitational acceleration. In other cases we may choose a different set of axes. Figure \(\PageIndex{1}\) illustrates the notation for displacement, where we define \(\vec{s}\) to be the total displacement, and \(\vec{x}\) and \(\vec{y}\) are its component vectors along the horizontal and vertical axes, respectively. The magnitudes of these vectors are s, x, and y.

An illustration of a soccer player kicking a ball. The soccer player’s foot is at the origin of an x y coordinate system. The trajectory of the soccer ball and its location at 6 instants in time are shown. The trajectory is a parabola. The vector s is the displacement from the origin to the final position of the soccer ball. Vector s and its x and y components form a right triangle, with s as the hypotenuse and an angle phi between the x axis and s.

To describe projectile motion completely, we must include velocity and acceleration, as well as displacement. We must find their components along the x- and y-axes. Let’s assume all forces except gravity (such as air resistance and friction, for example) are negligible. Defining the positive direction to be upward, the components of acceleration are then very simple:

\[a_{y} = -g = -9.8\; m/s^{2} (- 32\; ft/s^{2}) \ldotp\]

Because gravity is vertical, a x = 0. If a x = 0, this means the initial velocity in the x direction is equal to the final velocity in the x direction, or v x = v 0x . With these conditions on acceleration and velocity, we can write the kinematic Equation 4.11 through Equation 4.18 for motion in a uniform gravitational field, including the rest of the kinematic equations for a constant acceleration from Motion with Constant Acceleration. The kinematic equations for motion in a uniform gravitational field become kinematic equations with a y = -g, a x = 0:

Horizontal Motion

\[v_{0x} = v_{x}, \quad x = x_{0} + v_{x} t \label{4.19}\]

Vertical Motion

\[y = y_{0} + \frac{1}{2} (v_{0y} + v_{y})t \label{4.20}\]

\[v_{y} = v_{0y} - gt \label{4.21}\]

\[y = y_{0} + v_{0y} t - \frac{1}{2} g t^{2} \label{4.22}\]

\[v_{y}^{2}= v_{0y}^{2} + 2g(y - y_{0}) \label{4.23}\]

Using this set of equations, we can analyze projectile motion, keeping in mind some important points.

Problem-Solving Strategy: Projectile Motion

  • Resolve the motion into horizontal and vertical components along the x- and y-axes. The magnitudes of the components of displacement \(\vec{s}\) along these axes are x and y. The magnitudes of the components of velocity \(\vec{v}\) are v x = vcos\(\theta\) and v y = vsin\(\theta\), where v is the magnitude of the velocity and \(\theta\) is its direction relative to the horizontal, as shown in Figure \(\PageIndex{2}\).
  • Treat the motion as two independent one-dimensional motions: one horizontal and the other vertical. Use the kinematic equations for horizontal and vertical motion presented earlier.
  • Solve for the unknowns in the two separate motions: one horizontal and one vertical. Note that the only common variable between the motions is time t. The problem-solving procedures here are the same as those for one-dimensional kinematics and are illustrated in the following solved examples.
  • Recombine quantities in the horizontal and vertical directions to find the total displacement \(\vec{s}\) and velocity \(\vec{v}\). Solve for the magnitude and direction of the displacement and velocity using \(s = \sqrt{x^{2} + y^{2}} \ldotp \quad \phi = \tan^{-1} \left(\dfrac{y}{x}\right), \quad v = \sqrt{v_{x}^{2} + v_{y}^{2}} \ldotp\) where \(\phi\) is the direction of the displacement \(\vec{s}\).

Figure a shows the locations and velocities of a projectile on an x y coordinate system at 10 instants in time. When the projectile is at the origin, it has a velocity v sub 0 y which makes an angle theta sub 0 with the horizontal. The velocity is shown as a dark blue arrow, and its x and y components are shown as light blue arrow. The projectile’s position follows a downward-opening parabola, moving up to a maximum height, then back to y = 0, and continuing below the x axis The velocity, V, at each time makes an angle theta which changes in time, and has x component V sub x and y component v sub y. The x component of the velocity V sub x is the same at all times. The y component v sub y points up but gets smaller, until the projectile reaches the maximum height, where the velocity is horizontal and has no y component. After the maximum height, the velocity has a y component pointing down and growing larger. As the projectile reaches the same elevation on the way down as it had on the way up, its velocity is below the horizontal by the same angle theta as it was above the horizontal on the way up. In particular, when it comes back to y = 0 on the way down, the angle between the vector v and the horizontal is minus s=theta sub zero and the y component of the velocity is minus v sub 0 y. The last position shown is below the x axis, and the y component of the velocity is larger than it was initially. The graph clearly shows that the horizontal distances travelled in each of the time intervals are equal, while the vertical distances decrease on the way up and increase on the way down. Figure b shows the horizontal component, constant velocity. The horizontal positions and x components of the velocity of the projectile are shown along a horizontal line. The positions are evenly spaced, and the x components of the velocities are all the same, and point to the right. Figure c shows the vertical component, constant acceleration. The vertical positions and y components of the velocity of the projectile are shown along a vertical line. The positions are get closer together on the way up, then further apart on the way down. The y components of the velocities initially point up, decreasing in magnitude until there is no y component to the velocity at the maximum height. After the maximum height, the y components of the velocities point down and increase in magnitude. Figure d shows that putting the horizontal and vertical components of figures b and c together gives the total velocity at a point. The velocity V has an x component of V sub x, has y component of V sub y, and makes an angle of theta with the horizontal. In the example shown, the velocity has a downward y component.

Example \(\PageIndex{4}\): A Fireworks Projectile Explodes high and away

During a fireworks display, a shell is shot into the air with an initial speed of 70.0 m/s at an angle of 75.0° above the horizontal, as illustrated in Figure \(\PageIndex{3}\). The fuse is timed to ignite the shell just as it reaches its highest point above the ground. (a) Calculate the height at which the shell explodes. (b) How much time passes between the launch of the shell and the explosion? (c) What is the horizontal displacement of the shell when it explodes? (d) What is the total displacement from the point of launch to the highest point?

The trajectory of a fireworks shell from its launch to its highest point is shown as the left half of a downward-opening parabola in a graph of y as a function of x. The maximum height is h = 233 meters and its x displacement at that time is x = 125 meters. The initial velocity vector v sub 0 is up and to the right, tangent to the trajectory curve, and makes an angle of theta sub 0 equal to 75 degrees.

The motion can be broken into horizontal and vertical motions in which a x = 0 and a y = -g. We can then define x 0 and y 0 to be zero and solve for the desired quantities.

  • By “height” we mean the altitude or vertical position y above the starting point. The highest point in any trajectory, called the apex, is reached when v y = 0. Since we know the initial and final velocities, as well as the initial position, we use the following equation to find y: \(v_{y}^{2} = v_{0y}^{2} - 2g(y - y_{0}) \ldotp\) Because y 0 and v y are both zero, the equation simplifies to \(0 = v_{0y}^{2} - 2gy \ldotp\) Solving for y gives \(y = \frac{v_{0y}^{2}}{2g} \ldotp\) Now we must find v 0y , the component of the initial velocity in the y direction. It is given by v 0y = v 0 sin\(\theta_{0}\), where v 0 is the initial velocity of 70.0 m/s and \(\theta_{0}\) = 75° is the initial angle. Thus \(v_{0y} = v_{0} \sin \theta = (70.0\; m/s) \sin 75^{o} = 67.6\; m/s\) and y is \(y = \frac{(67.6\; m/s)^{2}}{2(9.80\; m/s^{2})} \ldotp\) Thus, we have \(y = 233\; m \ldotp\) Note that because up is positive, the initial vertical velocity is positive, as is the maximum height, but the acceleration resulting from gravity is negative. Note also that the maximum height depends only on the vertical component of the initial velocity, so that any projectile with a 67.6-m/s initial vertical component of velocity reaches a maximum height of 233 m (neglecting air resistance). The numbers in this example are reasonable for large fireworks displays, the shells of which do reach such heights before exploding. In practice, air resistance is not completely negligible, so the initial velocity would have to be somewhat larger than that given to reach the same height.
  • As in many physics problems, there is more than one way to solve for the time the projectile reaches its highest point. In this case, the easiest method is to use v y = v 0y - gt. Because v y = 0 at the apex, this equation reduces \(0 = v_{0y} - gt\) or \(t = \frac{v_{0y}}{g} = \frac{67.6\; m/s}{9.80\; m/s^{2}} = 6.90\; s \ldotp\) This time is also reasonable for large fireworks. If you are able to see the launch of fireworks, notice that several seconds pass before the shell explodes. Another way of finding the time is by using y = y 0 + \(\frac{1}{2}\)(v 0y + v y )t. This is left for you as an exercise to complete.
  • Because air resistance is negligible, a x = 0 and the horizontal velocity is constant, as discussed earlier. The horizontal displacement is the horizontal velocity multiplied by time as given by x = x 0 + v x t, where x 0 is equal to zero. Thus, \(x = v_{x} t,\) where v x is the x-component of the velocity, which is given by \(v_{x} = v_{0} \cos \theta = (70.0\; m/s) \cos 75^{o} = 18.1\; m/s \ldotp\) Time t for both motions is the same, so x is \(x = (18.1\; m/s)(6.90\; s) = 125\; m \ldotp\) Horizontal motion is a constant velocity in the absence of air resistance. The horizontal displacement found here could be useful in keeping the fireworks fragments from falling on spectators. When the shell explodes, air resistance has a major effect, and many fragments land directly below.
  • The horizontal and vertical components of the displacement were just calculated, so all that is needed here is to find the magnitude and direction of the displacement at the highest point: \(\vec{s} = 125 \hat{i} + 233 \hat{j}\) \(|\vec{s}| = \sqrt{125^{2} + 233^{2}} = 264\; m\) \(\theta = \tan^{-1} \left(\dfrac{233}{125}\right) = 61.8^{o} \ldotp\) Note that the angle for the displacement vector is less than the initial angle of launch. To see why this is, review Figure \(\PageIndex{1}\), which shows the curvature of the trajectory toward the ground level. When solving Example 4.7(a), the expression we found for y is valid for any projectile motion when air resistance is negligible. Call the maximum height y = h. Then, \(h = \frac{v_{0y}^{2}}{2g} \ldotp\) This equation defines the maximum height of a projectile above its launch position and it depends only on the vertical component of the initial velocity.

Example \(\PageIndex{5}\): Calculating projectile motion- Tennis Player

A tennis player wins a match at Arthur Ashe stadium and hits a ball into the stands at 30 m/s and at an angle 45° above the horizontal (Figure \(\PageIndex{4}\)). On its way down, the ball is caught by a spectator 10 m above the point where the ball was hit. (a) Calculate the time it takes the tennis ball to reach the spectator. (b) What are the magnitude and direction of the ball’s velocity at impact?

An illustration of a tennis ball launched into the stands. The player is to the left of the stands and hits the ball up and to the right at an angle of theta equal to 45 degrees and velocity of v sub 0 equal to 30 meters per second. The ball reaches a spectator who is seated 10 meters above the initial height of the ball.

Again, resolving this two-dimensional motion into two independent one-dimensional motions allows us to solve for the desired quantities. The time a projectile is in the air is governed by its vertical motion alone. Thus, we solve for t first. While the ball is rising and falling vertically, the horizontal motion continues at a constant velocity. This example asks for the final velocity. Thus, we recombine the vertical and horizontal results to obtain \(\vec{v}\) at final time t, determined in the first part of the example.

  • While the ball is in the air, it rises and then falls to a final position 10.0 m higher than its starting altitude. We can find the time for this by using Equation \ref{4.22}: \(y = y_{0} + v_{0y}t - \frac{1}{2} gt^{2} \ldotp\) If we take the initial position y 0 to be zero, then the final position is y = 10 m. The initial vertical velocity is the vertical component of the initial velocity: \(v_{0y} = v_{0} \sin \theta_{0} = (30.0\; m/s) \sin 45^{o} = 21.2\; m/s \ldotp\) Substituting into Equation \ref{4.22} for y gives us \(10.0\; m = (21.2\; m/s)t - (4.90\; m/s^{2})t^{2} \ldotp\) Rearranging terms gives a quadratic equation in t: \((4.90\; m/s^{2})t^{2} - (21.2\; m/s)t + 10.0\; m = 0 \ldotp\) Use of the quadratic formula yields t = 3.79 s and t = 0.54 s. Since the ball is at a height of 10 m at two times during its trajectory—once on the way up and once on the way down—we take the longer solution for the time it takes the ball to reach the spectator: \(t = 3.79\; s \ldotp\) The time for projectile motion is determined completely by the vertical motion. Thus, any projectile that has an initial vertical velocity of 21.2 m/s and lands 10.0 m above its starting altitude spends 3.79 s in the air.
  • We can find the final horizontal and vertical velocities v x and v y with the use of the result from (a). Then, we can combine them to find the magnitude of the total velocity vector \(\vec{v}\) and the angle \(\theta\) it makes with the horizontal. Since v x is constant, we can solve for it at any horizontal location. We choose the starting point because we know both the initial velocity and the initial angle. Therefore, \(v_{x} = v_{0} \cos \theta_{0} = (30\; m/s) \cos 45^{o} = 21.2\; m/s \ldotp\) The final vertical velocity is given by Equation \ref{4.21}: \(v_{y} = v_{0y} - gt \ldotp\) Since \(v_{0y}\) was found in part (a) to be 21.2 m/s, we have \(v_{y} = 21.2\; m/s - (9.8\; m/s^{2})(3.79 s) = -15.9\; m/s \ldotp\) The magnitude of the final velocity \(\vec{v}\) is \(v = \sqrt{v_{x}^{2} + v_{y}^{2}} = \sqrt{(21.2\; m/s)^{2} + (-15.9\; m/s)^{2}} = 26.5\; m/s \ldotp\) The direction \(\theta_{v}\) is found using the inverse tangent: \(\theta_{v} = \tan^{-1} \left(\dfrac{v_{y}}{v_{x}}\right) = \tan^{-1} \left(\dfrac{21.2}{-15.9}\right) = -53.1^{o} \ldotp\)
  • As mentioned earlier, the time for projectile motion is determined completely by the vertical motion. Thus, any projectile that has an initial vertical velocity of 21.2 m/s and lands 10.0 m above its starting altitude spends 3.79 s in the air.
  • The negative angle means the velocity is 53.1° below the horizontal at the point of impact. This result is consistent with the fact that the ball is impacting at a point on the other side of the apex of the trajectory and therefore has a negative y component of the velocity. The magnitude of the velocity is less than the magnitude of the initial velocity we expect since it is impacting 10.0 m above the launch elevation.

Example \(\PageIndex{5}\): Time of Flight for a Projectile on a Horizontal Surface

Solve for the time of flight of a projectile that is both launched and impacts on a flat horizontal surface.

We can solve for the time of flight of a projectile that is both launched and impacts on a flat horizontal surface by performing some manipulations of the kinematic equations. We note the position and displacement in y must be zero at launch and at impact on an even surface. Thus, we set the displacement in y equal to zero and find

\[y - y_{0} = v_{0y} t - \frac{1}{2} gt^{2} = (v_{0} \sin \theta_{0})t - \frac{1}{2} gt^{2} = 0 \ldotp\]

Factoring, we have

\[t \left(v_{0} \sin \theta_{0} - \dfrac{gt}{2}\right) = 0 \ldotp\]

Solving for t gives us

\[t_{tof} = \frac{2(v_{0} \sin \theta_{0})}{g} \ldotp \label{4.24}\]

This is the time of flight for a projectile both launched and impacting on a flat horizontal surface. Equation \ref{4.24} does not apply when the projectile lands at a different elevation than it was launched, as we saw in Example 4.8 of the tennis player hitting the ball into the stands. The other solution, t = 0, corresponds to the time at launch. The time of flight is linearly proportional to the initial velocity in the y direction and inversely proportional to g. Thus, on the Moon, where gravity is one-sixth that of Earth, a projectile launched with the same velocity as on Earth would be airborne six times as long.

Example \(\PageIndex{5}\): Trajectory of a Projectile on a Horizontal Surface

Find an expression for the trajectory of a projectile

The trajectory of a projectile can be found by eliminating the time variable t from the kinematic equations for arbitrary t and solving for y(x). We take x 0 = y 0 = 0 so the projectile is launched from the origin. The kinematic equation for x gives

\[x = v_{0x}t \Rightarrow t = \frac{x}{v_{0x}} = \frac{x}{v_{0} \cos \theta_{0}} \ldotp\]

Substituting the expression for t into the equation for the position y = (v 0 sin \(\theta_{0}\))t - \(\frac{1}{2}\) gt 2 gives

\[y = (v_{0} \sin \theta_{0}) \left(\dfrac{x}{v_{0} \cos \theta_{0}}\right) - \frac{1}{2} g \left(\dfrac{x}{v_{0} \cos \theta_{0}}\right)^{2} \ldotp\]

Rearranging terms, we have

\[y = (\tan \theta_{0})x - \Big[ \frac{g}{2(v_{0} \cos \theta_{0})^{2}} \Big] x^{2} \ldotp \label{4.25}\]

This trajectory equation is of the form y = ax + bx 2 , which is an equation of a parabola with coefficients

\[a = \tan \theta_{0}, \quad b = - \frac{g}{2(v_{0} \cos \theta_{0})^{2}} \ldotp\]

When we speak of the range of a projectile on level ground, we assume R is very small compared with the circumference of Earth. If, however, the range is large, Earth curves away below the projectile and the acceleration resulting from gravity changes direction along the path. The range is larger than predicted by the range equation given earlier because the projectile has farther to fall than it would on level ground, as shown in Figure \(\PageIndex{7}\), which is based on a drawing in Newton’s Principia . If the initial speed is great enough, the projectile goes into orbit. Earth’s surface drops 5 m every 8000 m. In 1 s an object falls 5 m without air resistance. Thus, if an object is given a horizontal velocity of 8000 m/s (or 18,000 mi/hr) near Earth’s surface, it will go into orbit around the planet because the surface continuously falls away from the object. This is roughly the speed of the Space Shuttle in a low Earth orbit when it was operational, or any satellite in a low Earth orbit. These and other aspects of orbital motion, such as Earth’s rotation, are covered in another chapter.

The figure shows a drawing of the earth with a tall tower at the north pole and a horizontal arrow labeled v 0 pointing to the right. 5 trajectories that start at the top of the tower are shown. The first reaches the earth near the tower. The second reaches the earth farther from the tower, and the third even farther. The fourth trajectory hits the earth at the equator, and is tangent to the surface at the equator. The fifth trajectory is a circle concentric with the earth.

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Free Fall Motion: Explanation, Review, and Examples

  • The Albert Team
  • Last Updated On: February 16, 2023

free fall problem with solution

Free fall and projectile motion describe objects that are moving through the air and acted on only by gravity. In this post, we will describe this type of motion using both graphs and kinematic equations. Since projectile motion involves two dimensions, these problems can be complex. We will explain many examples so you can see how to solve different types of projectile motion. 

What We Review

An object that is moving under only the influence of gravity is in free fall. In order for an object to be in free fall, wind and air resistance must be ignored. On Earth, all objects in free fall accelerate downward at the rate of gravity or 9.81\text{ m/s}^2 .

Applying Free Fall to Kinematic Equations

When analyzing free fall motion, we can apply the same kinematic equations as we did for motion on the ground. We can then use these equations to determine properties such as distance, time, and velocity. 

How to Find Distance Fallen for an Object in Free Fall

If an object is in free fall, we can use kinematic equations to find the distance it falls during a certain time. You will typically use the following kinematic equation to calculate the distance fallen:

In order to use this equation, you need to know the initial velocity of the object and the time of flight. Remember that the acceleration of a free falling object is always equal to the acceleration due to gravity, 9.81\text{ m/s}^2 . 

Many free fall physics problems will include scenarios where objects are dropped from rest. In this case, the initial velocity is zero and the first term of the kinematic equation above will cancel out. 

If the time is not known, another method for calculating the distance fallen is to use the following kinematic equation:

In this case, you must know the final velocity v_f of the object. Then, you can solve the equation for the distance d .

How to Find Time for an Object in Free Fall

The amount of time an object is in free fall will depend on its velocity and the distance it falls. Similar to distance, there are two equations you can use to find the time, depending on what you know. 

If you know the initial and final velocity of the object, then the simplest way to calculate time is using the kinematic equation:

This equation can be solved for time. Then, you’ll only need to substitute the values for the velocities and the acceleration due to gravity.

Another method to find time if you do not know the object’s final velocity is to use the equation:

Note that in this equation there are two terms that include the time t . Unless the initial velocity is zero, this can make it more challenging to solve this equation for time. If using this equation, you may need to use the quadratic formula to solve for time.

How to Find Final Velocity for an Object in Free Fall

The final velocity of an object in free fall depends on the amount of time it falls. Due to the acceleration of gravity, the velocity will increase every second by 9.81\text{ m/s} . The final velocity can be calculated using the equation:

If you do not know the amount of time the object is falling, another method for calculating the final velocity is using the kinematic equation: 

This equation requires that you instead know the distance that the object falls. If you are using this equation to find the final velocity, remember that the final velocity is squared in this equation. That means you will need to take a square root as your final step to solve for the final velocity. 

Examples of Free Fall

In this next section, we’ll apply the methods you just learned to solve some problems about free fall motion.

Example 1: How to Find the Distance for an Object Dropped from Rest

For example, an object is dropped from rest from the top of a tall building. It hits the ground 5\text{ s} after it is dropped. What is the height of the building? 

In this scenario, we know that the object’s initial velocity is zero because it was dropped from rest. We also know that the acceleration is 9.81\text{ m/s}^2 . This problem is asking us to find the distance the object falls. This will be equal to the height of the building.

Based on this information, we can use the following kinematic equation to find the distance:

Substituting the given values produces:

Therefore, the height of the building is about 123\text{ m} .

Example 2: How to Find the Final Velocity for an Object with Initial Velocity

In another example, an object in free fall has an initial, downward velocity of 2\text{ m/s} and falls a distance of 45\text{ m} . What is the object’s final velocity? 

In this scenario, we are given the object’s initial velocity, v_i and the distance d . We also know that the acceleration is 9.81\text{ m/s}^2 . Based on this information, we can use the following kinematic equation to find the final velocity:

Since the initial velocity is in the same direction as the acceleration (downward) we can use the same sign for both values.

Our last step is to eliminate the square by taking the square root:

Therefore, the final velocity of the object is about 30\text{ m/s} .

Motion Graphs for Objects in Free Fall

In addition to using physics equations, we can also represent free fall motion with motion graphs. Position-time graphs, velocity-time graphs, and acceleration-time graphs can tell us a lot about the object’s motion over time. Want a more in-depth review of motion graphs? Check out this blog post !

Position-Time Graph for an Object in Free Fall

In terms of position, many objects in free fall start at a high position, or height off the ground, and move downward. Objects in free fall accelerate due to gravity. Therefore, the position-time graph for free fall motion must be curved. This means that objects in free fall start with a slow velocity and gradually speed up which is represented by the steep downward curve of the graph. 

A position-time graph for an object in free fall will have a parabolic shape.

Velocity-Time Graph for an Object in Free Fall

As an object falls, its velocity increases due to the acceleration of gravity. This means that the velocity starts slow and steadily increases in the downward direction. The graph below shows the velocity-time for an object in free fall:

A velocity-time graph for an object in free fall will be a diagonal line with a negative slope.

Note that the slope of this graph is constant and represents the acceleration due to gravity, or -9.81\text{ m/s}^2 .

Acceleration-Time Graph for an Object in Free Fall

Free fall acceleration is constant. Throughout the entire time that an object is falling, it is accelerating at a rate equal to the acceleration due to gravity, -9.81\text{ m/s}^2 . As shown in the graph below, the acceleration-time graph is a constant negative line. 

An acceleration-time graph for an object in free fall will be a horizontal line with a constant value.

Projectile Motion

A projectile is an object that is launched or thrown into the air and then only influenced by gravity. Projectile motion has many similarities to free fall motion, however, projectiles may also travel a horizontal distance in addition to falling vertically down. 

Examples of Projectile Motion

The exact trajectory, or path, a projectile will take depends on how it is launched. However, all projectiles follow a curved trajectory such as in the image shown below:

The path of an object in projectile motion is called a trajectory and is a parabola.

If you play or watch sports, you likely have already observed projectile motion. Projectile motion describes the arc of a basketball in a free throw, a fly ball in baseball, or a volleyball bumped over the net. 

Horizontal Component of Velocity

To analyze projectile motion, we must separate the motion into horizontal and vertical components. The horizontal component of a projectile’s velocity is independent of the vertical component of velocity. Since gravity acts vertically, there are no horizontal forces acting on projectiles. This means that the horizontal component of a projectile’s velocity remains constant throughout the entire flight. 

Example: Finding the Horizontal Component

For example, a projectile is launched from the ground with an initial speed of 8\text{ m/s} at a 60^{\circ} angle. What is the horizontal component of the projectile’s velocity?

We will need to use trig identities to determine the components of the velocity. We can visualize the components as a triangle where the hypotenuse is the initial velocity and the sides represent the horizontal, v_{ix} , and vertical, v_{iy} , components of the velocity.

Objects experiencing projectile motion have a total velocity that can be analyzed as components using trig identities.

Cosine is defined as the adjacent side of the triangle divided by the hypotenuse. Since the horizontal component is adjacent to the angle, we can use cosine to find the horizontal component of velocity:

Therefore, the horizontal component of the initial velocity is 4\text{ m/s} .

Need to review your trig identities? Try out this resource from Khan Academy .

Vertical Component of Velocity

The vertical component of a projectile’s velocity will be influenced by gravity, which acts vertically on the object causing it to accelerate downward. Therefore, the vertical component of velocity will change throughout the projectile’s flight. We can calculate the vertical component of velocity at a particular time in a method similar to calculating the horizontal component. 

Example: Finding the Vertical Component

In the same example, a projectile is launched from the ground with an initial speed of 8\text{ m/s} at a 60^{\circ} angle. What is the vertical component of the projectile’s velocity?

As we visualize the velocity components, we are solving this time for the opposite side of the triangle. Sine is defined as the opposite side of the triangle divided by the hypotenuse. Therefore, the initial vertical velocity is:

Solving Projectile Motion Questions

Let’s apply what we’ve learned to some examples of projectile motion!

Example 1: Finding the Range of a Projectile

In this example, a projectile is fired horizontally with a speed of 5\text{ m/s} from a cliff with a height of 60\text{ m} . How far from the base of the cliff will the projectile land? 

In this scenario, we are given the initial horizontal velocity v_{ix}=5\text{ m/s} and the vertical change in position d_y=-60\text{ m} . Since the projectile is launched horizontally, the initial vertical velocity, v_{iy} , is zero. We also always know in projectile motion that the vertical acceleration is a_y=-9.81\text{ m/s}^2 and the horizontal acceleration, a_x , is zero.

This problem is asking us to find the horizontal displacement, or d_x . This is also referred to as the range . We can use the following kinematic equation to find the projectile’s final horizontal position:

Since the horizontal acceleration of a projectile is zero, this equation can be simplified to:

Before we can solve this equation, we must first determine the time of the projectile’s flight. We can actually use this same equation in the vertical direction to solve for time:

Since the initial vertical velocity is zero, this equation can be simplified to:

Solving for t :

Substituting the given values:

Now we can use this time to calculate the horizontal displacement of the projectile:

Therefore, the projectile will land about 17.5\text{ m} from the base of the cliff. 

Example 2: Finding the Maximum Height of a Projectile

As another example, a projectile is launched from the ground with an initial velocity of 25\text{ m/s} at an angle of 50^{\circ} . What is the projectile’s maximum height?

As a projectile travels upward, its vertical velocity becomes slower and slower due to the negative acceleration of gravity. At the maximum height of the trajectory, the projectile’s vertical velocity will momentarily be zero as the projectile stops and turns to move downward. Therefore, in this scenario, our final vertical velocity, v_{fy} , is zero.

We can use the following kinematic equation to solve for the maximum height, d_y :

Solving for d_y :

Before we can use this equation to calculate the height, we will need to use the sine trig identity to find the vertical component of the initial velocity:

Since the initial velocity is in the opposite direction as the acceleration, it’s really important to remember the sign here. If we define moving up as positive, then the initial velocity is positive and the acceleration is negative. Substituting this initial vertical velocity and the given values into the equation above gives:

Therefore, the projectile will reach a maximum height of about 18.7\text{ m} .

For more examples and an explanation of solving these types of projectile motion problems, check out this youtube video from Professor Dave . 

Understanding free fall and projectile motion allows you to solve some of the most complex problems you will encounter in introductory physics. All projectiles are acted on only by gravity, and the vertical and horizontal components of motion are independent of each other. This allows us to apply our kinematic equations to solve for a projectile’s time of flight, velocity, and displacement in each direction.  

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Free fall

Look at the given example below and try to understand what I tried to explain above.

Free fall

Kinematics Exams and Solutions

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Free fall motion – problems and solutions

1. A stone free fall from the height of 45 meters. If the acceleration due to gravity is 10 ms -2 , what is the speed of the stone when hits the ground?

Height (h) = 45 meters

Acceleration due to gravity (g) = 10 m/s 2

Wanted : The final velocity of the stone when it hits the ground (v t )

The equation of free fall motion :

v t 2 = 2 g h

The final velocity of the stone :

v t 2 = 2 (10)(45) = 900

v t = √900 = 30 m/s 2

2. An object free fall from a height without the initial velocity. The object hits the ground 2 seconds later. Acceleration due to gravity is 10 ms -2 . Determine height

Time interval (t) = 2 seconds

Wanted : Height (h)

The equation of the free fall motion :

h = ½ g t 2

h = ½ (10)(2) 2 = (5)(4) = 20 meters

3.A 2-kg object free fall from a height of 20 meters above the ground. What is the time interval the object in air ? Acceleration due to gravity is 10 ms -2

Height (h) = 20 meters

Wanted : Time interval (t)

Time interval :

20 = ½ (10)(t 2 )

20 = (5)(t 2 )

t = 2 seconds

4. Two objects, object 1 and object 2, are free fall from a height of h 1 and h 2 at the same time. If h 1 : h 2 = 2: 1, what is the ratio of the time interval of the object 1 to the object 2.

The height of the object 1 (h 1 ) = 2

The height of the object 2 (h 2 ) = 1

Acceleration due to gravity = g

Wanted : t 1 : t 2

h 1 = 1/2 g t 1 2

2 = 1/2 g t 1 2

(2)(2) = g t 1 2

4 = g t 1 2

4/g = t 1 2

h 2 = 1/2 g t 2 2

1 = 1/2 g t 2 2

(2)(1) = g t 2 2

2 = g t 2 2

2/g = t 2 2

The ratio of the time interval :

√4/g : √2/g

(√4/g) 2 : (√2/g) 2

5. An object dropped from a height of h above the ground. The final velocity when object hits the ground is 10 m/s. What is the time interval to reach ½ h above the ground. Acceleration due to gravity is 10 m/s 2 .

The final velocity (v t ) = 10 m/s

Wanted : The time interval to reach 1/2 h above the ground

The height of h :

10 2 = 2 (10) h

h = 100 / 20

h = 5 meters

The height of 1/2 h = 1/2 (5 meters) = 2.5 meters. The time interval needed to reach 2.5 meters above the ground :

h = 1/2 g t 2

2.5 = 1/2 (10) t 2

2.5 = 5 t 2

t 2 = 2.5 / 5 = 0.5 = (0.25)(2)

t = √(0.25)(2) = 0.5√2 = 1/2 √2 seconds

Free fall motion – problems and solutions 1

T he free fall motion of coconut ( figure 1 ) and the motion of a ball thrown vertically up ward to the highest poin t by a student (figur e 2). Determine the kind of both motion s.

Free fall motion – problems and solutions 2

Figure 1 = free fall motion = Acceleration

Figure 2 = vertical motion = Deceleration

The correct answer is A.

7. A stone free fall from a building. The time interval needed by a stone to reach the ground is 3 seconds and acceleration due to gravity is 10 m/s 2 . Determine the height of the building.

Time interval (t) = 3 seconds

Acceleration due to gravity (g) = 10 m.s -2

Wanted: Height of building (h)

Known: time interval (t) and acceleration due to gravity (g), wanted: height (h) so use the equation of free fall motion: h = ½ g t 2

h = ½ (10)(3)

h = 15 metes

8. A fruit free fall from its tree at the height of 12 m above the ground. If acceleration due to gravity is g = 10 m/s 2 and the friction of air ignored, then determine the height of the fruit above the ground after 1 second.

Height of tree (h) = 12 meters

Time interval (t) = 1 second

Wanted: The height of the fruit above the ground

After 1 second, fruit free fall as far as :

h = ½ g t 2 = ½ (10)(1) 2 = (5)(1) = 5 meters

The height of the fruit above the ground after 1 second :

12 meters – 5 meters = 7 meters

Answer : Free fall motion refers to the motion of an object under the influence of gravitational force only, with no other forces (like air resistance) acting on it.

Answer : All objects in free fall near the surface of the Earth experience a constant acceleration due to gravity, g , which is approximately 9.81 m/s 2 downward. This means that the object’s velocity increases by this amount for each second of free fall.

Answer : In the absence of air resistance, the mass of an object does not influence its free fall acceleration. All objects, regardless of their mass, will fall with the same acceleration due to gravity, g .

Answer : Astronauts inside the ISS appear to float not because there’s no gravity, but because both the astronauts and the ISS are in a continuous state of free fall around the Earth. They’re essentially falling at the same rate as the ISS, creating a sensation of weightlessness.

Answer : Mass is a measure of the amount of matter in an object and remains constant regardless of its location. Weight, on the other hand, is the force exerted on an object due to gravity. It varies depending on the gravitational field. During free fall, an object feels weightless because there’s no normal force acting on it, but its mass remains unchanged.

Answer : When an object is thrown upwards, it decelerates under the influence of gravity. Its velocity decreases until it becomes zero at its highest point. As it starts falling back down, it accelerates due to gravity, increasing its velocity in the downward direction.

Answer : Terminal velocity is the constant maximum velocity reached by a falling object when the downward force of gravity is balanced by the upward force of air resistance. At this point, the object no longer accelerates and continues to fall at a constant speed.

Answer : The time it takes for an object to reach the ground is proportional to the square root of the height from which it falls (assuming no air resistance). An object dropped from a greater height will take longer to reach the ground than one dropped from a shorter height.

Answer : As an object falls freely under gravity, its potential energy (relative to the ground) decreases. This decrease in potential energy is converted into kinetic energy, causing the object’s speed to increase.

  • If two objects of different shapes but the same mass are dropped from the same height in a vacuum, which will hit the ground first?

Answer : In a vacuum, where there’s no air resistance, both objects will hit the ground at the same time. Their shape won’t matter because only gravity is acting on them, and their masses are the same.

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practice problem 1

An hour and thirty-one minutes after launch, my pressure altimeter halts at 103,300 feet. At ground control the radar altimeters also have stopped on readings of 102,800 feet, the figure that we later agree upon as the more reliable. It is 7 o'clock in the morning, and I have reached float altitude. At zero count I step into space. No wind whistles or billows my clothing. I have absolutely no sensation of the increasing speed with which I fall. Though my stabilization chute opens at 96,000 feet, I accelerate for 6,000 feet more before hitting a peak of 614 miles an hour, nine-tenths the speed of sound at my altitude. An Air Force camera on the gondola took this photograph when the cotton clouds still lay 80,000 feet below. At 21,000 feet they rushed up so chillingly that I had to remind myself they were vapor and not solid. Joseph Kittinger, 1960

For most skydivers, the acceleration experienced while falling is not constant. As a skydiver's speed increases, so too does the aerodynamic drag until their speed levels out at a typical terminal velocity of 55 m/s (120 mph). Air resistance is not negligible in such circumstances. The story of Captain Kittinger is an exceptional one, however. At the float altitude where his dive began, the Earth's atmosphere has only 1.5% of its density at sea level. It is effectively a vacuum and offers no resistance to a person falling from rest.

The acceleration due to gravity is often said to be constant, with a value of 9.8 m/s 2 . Over the entire surface of the Earth up to an altitude of 18 km, this is the value accurate to two significant digits. In actuality, this "constant" varies from 9.81 m/s 2 at sea level to 9.75 m/s 2 at 18 km. At the altitude of Captain Kittinger's dive, the acceleration due to gravity was closer to 9.72 m/s 2 .

Given this data it is possible to calculate the maximum speed of Captain Kittinger during his descent. First we will need to convert the altitude measurements. To save calculation time we will only convert the change in altitude and not each altitude. Given that he stepped out of the gondola at 102,800 feet, fell freely until 96,000 feet, and then continued to accelerate for another 6,000 feet; the distance over which he accelerated uniformly was…

It's now just a matter of choosing the correct formula and plugging in the numbers.

This result is amazingly close to the value recorded in Kittinger's report.

As one would expect the actual value is slightly less than the theoretical value. This agrees with the notion of a small but still non-zero amount of drag.

practice problem 2

  • just before it hit the floor on the way down
  • just after it left the floor on the way up
  • on the way down
  • while it is contact with the floor
  • on the way up

The first half of this problem is much like every other falling body problem. Dropped basketballs speed up on the way down, so acceleration is positive.

The second half of this problem is similar to the first half, only the velocity reduces to zero instead of starting at zero. This means the acceleration should probably be negative.

There is another way to solve the second half of this question using the notion of proportionality. 0.67 m is approximately ⅔ of 1.00 m. Velocity is proportional to the square root of distance when acceleration is constant (and the initial velocity is zero).

v  ∝ √∆ s

This method is no easier, but it serves as a check on our first calculation. Since both results are identical, we've probably done this right.

This question is designed to see if you've been paying attention. The acceleration of a freely falling body is 9.8 m/s 2  down near the surface of the Earth.

This part requires computation. Use the definition of acceleration. Let's say that down is negative. Then…

There is little work to do here. Just write the answer. The acceleration due to gravity is still 9.8 m/s 2  down even if the basketball is rising.

practice problem 3

  • the time the diver was in the air
  • the maximum height to which she ascended
  • her velocity on impact with the water

This is a problem in which close attention to signs is a must. Let's assume that up is positive. Start with the givens and the unknown.

Pick an appropriate equation and substitute.

We have a quadratic here, so we reach for the quadratic equation. You'll excuse me if I drop the units for a moment. We need to see the numbers clearly.

Quadratics have two solutions. What do they mean? Surely the negative solution is nonsense. How could a diver dive and yet strike the water before she left the board? The answer is 1.5 s .

At the point of maximum height, the diver's velocity would be zero. Thus…

Select a good equation and substitute.

This is the height above the board. To get the height above the water, add the height of the board.

Solving for the impact velocity is perhaps the easiest problem. Just get the direction right.

practice problem 4

  • displacement-time
  • velocity-time
  • acceleration-time

The displacement-time graph is the easiest for most people to think about. The object goes up for a while and then comes down. Since the velocity is changing, the graph is curved. Looking at the slope of the tangent to the curve, we can see that the object starts with a positive (upward) velocity that decreases to zero at the top where the graph turns around. The slope then turns negative and gets steeper as the velocity increases downward.

The velocity-time graph is trickier since many people can't separate it mentally from the previous graph. Using the slope of the tangent of the displacement-time graph can help us. The object's velocity starts out large and positive, decreases at a uniform rate until it reaches zero, then keeps decreasing (or gets more negative, whichever you prefer). At the end, the object is moving downward just as fast as it was moving upward at the beginning.

The acceleration due to gravity is constant and directed downward. It neither increases, nor decreases, nor changes in any significant way. Looking at the previous graph, this should be apparent. The velocity of an object grows in the negative (downward) direction at a constant rate. When a value is constant, its graph with respect to time is a horizontal line.

Guy jumps from the 3rd floor onto a tree. Wild stuff

Galileo (1564 - 1642)

free fall problem with solution

Credit : Original portrait by Justus Sustermans painted in 1636.

[Flash] Rutgers U.

Click on the Arrow

Free Fall - D vs. t Plot Rutgers U.

Distance and Time Rutgers U.

(Roll your mouse over this)

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Gravity/Moon - Feather & Hammer NASA

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g = - 9.8 m/s 2

Rounding off g we get -10 m/s 2

Ex) An object thrown upward at +40 m/s

An object is thrown up at a velocity of 30 m/s

About how long will it take to hit the ground?

Draw the V vs t Plot

free fall problem with solution

A. Motion diagram for free fall on any planet or moon

Label the slope pictured above that could be Jupiter? Moon?

 Top Line: Jupiter

Bottom Line: Moon 

D vs t - free fall

free fall problem with solution

A ball is thrown upward. What's the acceleration  due to gravity at the peak?

AP Question

Ex) Compare the displacement of an object dropped during its first second with its  displacement after 3 seconds.

 Δd = V i t + 1/2aΔt 2

 Δd = 1/2aΔt 2

9 times greater

Effect of Mass on Falling Bodies

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Free Fall Practice Problems for High Schools: Complete Guide

In this long article, we are going to practice some problems about a freely falling object in the absence of air resistance. All these questions are suitable for high school or college students, or even the AP Physics 1 exam. 

Freely Falling Motion Problems

Problem (1): A tennis ball is thrown vertically upward with an initial speed of 17 m/s and caught at the same level above the ground.  (a) How high does the ball rise? (b) How long was the ball in the air? (c) How long does it take to reach its highest point?

Solution : Take up as the positive direction and the throwing point as the origin, so $y_0=0$. 

(a) The ball goes up so high that its vertical velocity becomes zero. For this part of ascending motion, we can use the free fall kinematic equation $v^2-v_0^2=-2g(y-y_0)$. Substituting the known values into it and solving for $y$, we get \begin{gather*} v^2-v_0^2=-2g(y-y_0) \\\\ 0-17^2 = -2(10)(y_{max}-0) \\\\ \Rightarrow \boxed{y_{max}=14.45\,\rm m} \end{gather*}  (b) In all free fall practice problems, the best way to find the total flight time that the object was in the air is to use the kinematic equation $y-y_0=-\frac 12 gt^2+v_0t$. Then, substitute the coordinates of where the object landed on the ground into it. 

As a rule of thumb, if the object returns to the same point of launch, its displacement vector is always zero, so $y-y_0=0$. Therefore, we have \begin{gather*} y-y_0=-\frac 12 gt^2+v_0t \\\\ 0=-\frac 12 (10)t^2+17t \\\\ \Rightarrow \boxed{5t^2-17t=0} \end{gather*} Solving this equation by factoring out the time and setting the remaining expression to zero, we get  \begin{gather*} 5t^2-17t=0 \\\\ t(5t-17)=0 \\\\ \Rightarrow t=0 \quad ,\quad t=3.4\,\rm s \end{gather*} The first result corresponds to the initial time, and the other time, $\boxed{t_{tot}=3.4\,\rm s}$, is the amount of time the ball is in the air until it reaches the ground. 

(c) At the highest point the vertical velocity is always zero, $v=0$. Using the equation $v=v_0-gt$, and solving for $t$, we have \begin{gather*} v=v_0-gt \\ 0=17-(10)t \\ \Rightarrow t_{top}=1.7\,\rm s \end{gather*} As you can see, the duration of the ball's going up, in the absence of air resistance, is always half the total flight time. Therefore, \[t_{top}=\frac 12 t_{tot}\]

Problem (2): A ball is dropped directly downward from a height of 45 meters with an initial speed of 6 m/s. How many seconds later does it strike the ground?

Solution : Taking up as the positive direction and the dropping point as the origin, we have $y_0=0$. Since the ball is moving downward, we choose a negative sign for its initial velocity, so $v_0=-6\,\rm m/s$.

The ball strikes the ground $45\,\rm m$ below the chosen origin, so its correct coordinate is $y=-45\,\rm m$. 

The only kinematic equation that relates all these variables to the time is $y-y_0=-\frac 12 gt^2+v_0t$. Substituting the numerical values into this equation, yields \begin{gather*} y-y_0 =-\frac 12 gt^2+v_0t \\\\ -45-0 =-\frac 12 (10)t^2+(-6)t \\\\ \Rightarrow \boxed{5t^2+6t-45=0} \end{gather*} In the last step, after rearranging, we arrived at a quadratic equation, like $at^2+bt+c=0$, that its solution is found using the below formula \[t=\frac{-b\pm\sqrt{b^2-4ac}}{2a} \] where $a,b,c$ are constants. In this case, we have \[a=5, b=5, c=-45\] Substituting these values into the above formula, we get \begin{gather*} t=\frac{-5\pm\sqrt{5^2-4(5)(-45)}}{2(5)} \\\\ t=2.46\,\rm s \quad , \quad t=-3.66 \end{gather*} Since we choose the positive time for free fall problems, the ball would reach the ground approximately $2.5\,\rm s$ after being dropped.

Problem (3): A coin is tossed vertically upward and remains $\rm 0.6\, s$ in the air before it comes back down.  (a) How fast is the coin tossed? (b) How high does the coin rise? 

Solution : Let the tossing point be the origin, so $y_0=0$. The time duration the coin is in the air until it reaches the highest point is $t=0.6\,\rm s$. Recall that in all free fall problems, the object at the highest point has a velocity $v=0$. 

(a) Use the kinematic equation $v=v_0-gt$, substitute the above-known values, and solve for the unknown initial speed $v_0$ \begin{align*} v&=v_0-gt \\ 0&=v_0-(10)(0.6) \\ \Rightarrow v_0&=6\,\rm m/s \end{align*} Therefore, the coin was tossed vertically upward with an initial speed of $6\,\rm m/s$. 

(b) In this part, we need a kinematic equation that relates the distance traveled to the time taken, $y-y_0=-\frac 12 gt^2+v_0t$, or to the initial and final velocities, $v^2-v_0^2=-2g(y-y_0)$. We want to use the second equation as follows. \begin{align*} v^2-v_0^2&=-2g(y-y_0) \\\\ (0)^2-(60)^2 &= -2(10)(y-0) \\\\ y&=\frac{-6^2}{-20} \\\\ &=\boxed{1.8\,\rm m} \end{align*} You can also use the first equation and arrive at the same result. Check it out yourself. 

Problem (4): A small stone is shot straight up with an initial speed of $\rm 15\,m/s$.  (a) How long does the stone take to reach its maximum height? (b) How high does the stone go? (c) With what speed would the stone hit the ground?

Solution : The known data is $y_0=0$ and $v_0=15\,\rm m/s$. 

(a) When an object is thrown vertically upward in the air, it rises until it reaches a point where its vertical velocity becomes zero; otherwise, it continues on its way. So, in this part, at the maximum height, we have $v=0$. Substituting all these known numerical values into the equation $v=v_0-gt$, and solving for the unknown time $t$, we get \begin{align*} 0 &= 15-(10)t \\ \Rightarrow t&=\boxed{1.5\,\rm s} \end{align*} Thus, it takes $1.5\,\rm s$ for  the stone to reach the highest point. 

(b) For this part, we have the time taken to reach the maximum height $t=1.5\,\rm s$, the initial and final velocities. Thus, we can use either the equation $y-y_0=-\frac 12 gt^2+v_0t$ or $v^2-v_0^2=-2g(y-y_0)$. We chose the second equation. \begin{align*} 0-(15)^2 &= -2(10)(y_{max}-0) \\\\ y_{max} &=\frac{-15^2}{-20} \\\\ \Rightarrow y_{max}&= \boxed{11.25\,\rm m} \end{align*} Thus, the stone reaches the maximum height of $11.25\,\rm m$. 

(c) When the stone hits the ground at the same level as the throwing point, then according to the definition of displacement, it displaces nothing. This means that the displacement between initial and final points $\Delta y=y-y_0$ is zero. 

Using this fact, we can use the equation $v^2-v_0^2=-2g(y-y_0)$ to get the velocity at the moment of hitting the ground. \begin{align*} v^2-(15)^2 &=-2(10)(0) \\ \Rightarrow v^2 = 15^2 \end{align*} Taking the square root of both sides, yields two roots of $v=\pm 15\,\rm m/s$. Recall that velocity is a vector in physics that has both magnitude and direction. 

The stone is hitting the ground, so the correct sign is a negative for its velocity, i.e., $v=-15\,\rm m/s$. 

Problem (5): From the top of a high cliff, a stone is released. It is seen after $2.5\,\rm s$, the stone strikes the ground. How high is the cliff? 

Solution : The stone is released, so its initial speed is zero, $v_0=0$. The total flight time is also $t=2.5\,\rm s$. Taking the releasing point as the origin, we will have $y_0=0$. In the kinematic equations, there is vertical displacement in only two of them, i.e., $v^2-v_0^2=-2g(y-y_0)$ and $y-y_0=-\frac 12 gt^2+v_0t$. As you can see, to use the first equation, we must have the final velocity where the stone hits the ground, and for the second equation, the total flight time is needed. Hence, it is simpler to apply the second equation and solve for the unknown vertical distance $y$.  \begin{gather*} y-y_0=-\frac 12 gt^2+v_0t \\\\ y-0=-\frac 12 (10)(2.5)^2+(0)(2.5) \\\\ \Rightarrow \quad \boxed{y=-31.25\,\rm m} \end{gather*} The negative indicates that the stone strikes the ground $31.25\,\rm m$ below our chosen origin. 

Problem (6): A stone is released at rest from a height and falls freely for $4\,\rm s$.  (a) What is the stone's velocity $1.5\,\rm s$ after releasing? (b) How high is the height?

Solution : The stone is released at rest, so $v_0=0$. Take upward as the positive direction. The total flight time is $t_{tot}=4\,\rm s$. 

(a) With this known information, use the equation $v=v_0-gt$ to find the stone's velocity at each instant of time. \begin{gather*} v=v_0-gt \\ v=0-(10)(1.5) \\ \Rightarrow \boxed{v=-15\,\rm m/s} \end{gather*} The negative indicates the direction of the stone at that moment, which is facing down. 

(b) The time between releasing the stone and hitting the ground is given. With this known information, use the equation $y-y_0=-\frac 12 gt^2 +v_0t$ and solve for the total distance fallen by the stone. \begin{gather*} y-0=-\frac 12 (10)(4)^2+(0)(4) \\\\ \Rightarrow \quad \boxed {y=-80\,\rm m} \end{gather*} The minus sign reminds us that the stone strikes the ground $80\,\rm m$ below our chosen origin. 

Problem (7): A person throws a light stone straight up and catches it $2.6\,\rm s$ later. With what speed did he throw the stone, and to what height does the stone go up? 

Solution : As with any other free-fall problem, let upward be the positive direction and the throwing point be the origin of the coordinate system so that $y_0=0$. 

The stone is caught at the same level of throwing, so its vertical displacement is zero, $\Delta y=y-y_0=0$. The total flight time is also known. Hence, applying the vertical displacement kinematic equation, $y-y_0=-\frac 12 gt^2+v_0t$, and solving for the initial speed $v_0$ gives us \begin{gather*} y-y_0=-\frac 12 gt^2+v_0t \\\\ 0=-\frac 12 (10)(2.6)^2+v_0 (2.6) \end{gather*} Factoring out $2.6$ from the last expression, would get \begin{gather*} (2.6)(-5(2.6)+v_0)=0 \\\\ \Rightarrow \quad \boxed{v_0=13\,\rm m/s} \end{gather*} As a side note, if an object is thrown vertically upward and caught at the same level, by knowing the time interval between these two moments, we can use the following formula to find its initial speed \[v_0=\frac 12 g t\] The stone goes up until it reaches a point where its vertical velocity is zero, i.e., $v=0$. Now that we know the stone's initial speed, applying the equation $v^2-v_0^2=-2g(y-y_0)$ gives us \begin{gather*} 0-(13)^2=-2(10)(y-0) \\\\ \Rightarrow \quad \boxed{y=8.45\,\rm m} \end{gather*} 

Problem (8): A baseball is thrown straight up with a speed of $25\,\rm m/s$.  (a) With what speed is it moving when it is at a height of $10\,\rm m$? (b) How much time does it take to reach that point?

Solution : As usual, take up the positive $y$-direction and set $y_0=0$. The initial speed is also $v_0=25\,\rm m/s$. 

(a) The only time-independent kinematic equation that relates these known values to each other is $v^2-v_0^2=-2g(y-y_0)$. Substituting the known numerical values into this, we get \begin{gather*} v^2-(25)^2=-2(9.8)(10-0) \\\\ v^2=429 \\\\ \Rightarrow \quad v=\pm 20.7\,\rm m/s \end{gather*} As you can see, we obtained a speed with two different signs. Here, the speed with a positive sign indicates that the baseball at that desired height is being moved upward, while the negative sign hints to us that the baseball is at that height when it is moving down. 

Thus, we arrive at the fact that the baseball reaches that height twice, once when it is going up and once when it is going down.

(b) In the previous part, we found out that we were at that height twice. Thus, there are two corresponding times for this situation. Applying the equation $y-y_0=-\frac 12 gt^2+v_0t$ and solving for the required time $t$, we get \begin{gather*} 10-0=-\frac 12 (10)t^2+25t \\\\ \Rightarrow \quad 4.9t^2-25t+10=0 \end{gather*} The quadratic equation above has two solutions as below \begin{gather*} t_1=0.44\,\rm s \quad , \quad t_2=4.66\,\rm s \end{gather*} the first time, $t_1$ is when the ball is going up, and the second one corresponds to when it is moving down. 

Note : The quadratic equation of $at^2+bt+c=0$ has the following solution formula \[t=\frac{-b\pm\sqrt{b^2-4ac}}{2a}\]

Problem (9): A helicopter is ascending vertically upward with a constant speed of $6.5\,\rm m/s$ and carrying a package of $2\,\rm kg$. When it reaches a height of $150\,\rm m$ above the surface, it drops the package. How much time does it take for the package to hit the surface?

Solution : Take up the positive direction, as always. At the moment that the package is released, it has the speed of its carrier. 

There is a subtle point in this case. Since the object is moving down, in the opposite direction of our positive direction, a negative must accompany it. Therefore, the correct input for the initial speed is $v_0=-6.5\,\rm m/s$.

Applying the equation $y-y_0=-\frac 12 gt^2+v_0t$, we will have \begin{gather*} y-y_0=-\frac 12 gt^2+v_0t \\\\ -150-0=-\frac 12 (10)t^2+(-6.5)t \\\\ \Rightarrow 5t^2+6.5t-150=0 \end{gather*} Using a graphing calculator or the formula in the previous question, we find that it takes about $4.9\,\rm s$ for the package to reach the ground. 

Note that in the above, we inserted a negative for the vertical height as $y=-150\,\rm m$, since the package hits the ground $150\,\rm m$ below our chosen origin. 

Problem (10): A stone is thrown vertically upward from a building $15\,{\rm m}$ high with an initial velocity of $10\,{\rm m/s}$. What is the stone's velocity just before hitting the ground?

Solution : In all freely falling practice problems, the important note is choosing the origin. Usually, the throwing (releasing or dropping) point is the best choice. In this case, the hitting point is below the origin, so its vertical displacement ($y$) is negative.

Applying the time-independent free fall kinematic equation, we have \begin{align*}v_f^{2}-v_i^{2}&=2\,(-g)\Delta y\\v_f^{2}-(10)^{2}&=2\,(-10)(-15) \\ \Rightarrow v_f&=\pm 20\,{\rm m/s}\end{align*} Since velocity is a vector quantity and just before striking the ground its direction is vertically downward, the negative value must be chosen, i.e., $v_f=-20\,{\rm m/s}$. 

Problem (11): A bullet is fired vertically downward with an initial speed of $15\,{\rm m/s}$ from the top of a tower of $20\,{\rm m}$ high. What is its velocity at the instant of striking the ground? 

Solution : Let the origin be the firing point. Using the below kinematic equation, we have \begin{align*}v_f^{2}-v_i^{2}&=-2g(y-y_0) \\\\ v_f^{2}-0&=-2(10)(-20)\\\Rightarrow v_f&=\pm 20\,{\rm m/s}\end{align*} Since the direction of velocity at the moment of striking the ground is downward, we must choose the negative correct sign, so we have $v_f=-20\,\rm m/s$. 

Problem (12): There is a well with a depth of $34\,{\rm m}$. A person drops a stone vertically into it with an initial velocity of $7\,{\rm m/s}$. What is the time interval between dropping the stone and hearing its impact sound? (Assume $g=10\,{\rm m/s^2}$ and the speed of the sound in the air is $340\,{\rm m/s}$).

Solution : This motion has two parts. One is descending into the well, which is a constant acceleration motion, and the other is ascending to the sound of impact, which is a uniform motion with constant speed. 

For the first part, use the kinematic equation $y-y_0=-\frac 12 gt^{2}+v_0 t$ to find the falling time as \begin{gather*}y-y_0=-\frac 12 gt^{2}+v_0 t \\\\ -34-0=-\frac 12 (10)t^{2}+(-7)t \\\\ \Rightarrow 5t^{2}+7t-34=0 \end{gather*} Since initial velocity is a vector and its direction is initially downward, a negative is included in front of it. The above quadratic equation has two solutions: $t_1=2\,{\rm s}$ and $t_2=-3.4\,{\rm s}$. Obviously, a negative value for time is not accepted because time in all kinematic questions must be a positive quantity.

The second part is a uniform motion as the speed of sound is constant, so we can calculate its rising time using the definition of average velocity: \begin{align*} t&=\frac{\Delta y}{v} \\\\ &=\frac{34}{340}=0.1\,{\rm s}\end{align*} Thus, the total time is obtained as $t_T=2+0.1=2.1\,{\rm s}$. 

Problem (13): A stone is dropped vertically downward from the top of a building with a height of $h$. What is its speed at the height of $\frac h2$?

Solution : Let the dropping point be the origin. Thus, in the kinematic equations, the vertical displacement must be negative, i.e., $y-y_0=-\frac h2$. Use the below equation to find the speed at the desired level \begin{align*}v_2^2-v_1^2&=-2g(y-y_0) \\\\ v_2^2 -0&=-2g(-\frac h2) \\\\ \Rightarrow v_2&=\sqrt{gh}\end{align*}

Problem (14): You throw a baseball straight up in the air and catch it 3.4 seconds later at the same place at which you threw it. How high did it go? (DIFFICULT)

Solution : First of all, choose a coordinate system with the upward direction as positive. In the absence of air resistance, when you throw an object upward, the time it takes to rise is equal to the time it takes to fall. In other words, the time of reaching the highest point is half the total time the object is in the air \[t_{rise}=t_{fall}=\frac 12 t_{tot}\] Other than the total time of flight, no other relevant information is given. Substitute the time taken to reach the highest point into $v=v_0-gt$ and solve for the initial velocity $v_0$ as below \begin{gather*} v_{top}=v_0-gt_{rise} \\\\ 0=v_0-(9.8)(1.7) \\\\ \Rightarrow v_0=16.6\,\rm m/s\end{gather*} Note that at the highest point $v_{top}=0$. 

Now, apply the equation $v^2-v_0^2=-2g\Delta y$ between the initial point and the highest point so that $\Delta y=h$ and solve for $h$. \begin{gather*} v^2_{top}-v_0^2=-2gh \\\\ (0)^2-(16.6)^2=-2(9.8)h \\\\ \Rightarrow \boxed{h=14.0\,\rm m} \end{gather*} Therefore, the ball rises as high as $14.0\,\rm m$.

Problem (15): A bullet is fired vertically upward from a height of $90\,{\rm m}$ and after $10\,{\rm s}$ reaches the ground. After $2\,{\rm s}$ from the throwing point, the bullet is how far away from the surface? ($g=9.8\,{\rm m/s^2}$)

Solution : In a free-fall problem, the vertical position of an object in an instant of time is given by the kinematic equation $y=-\frac 12 gt^{2}+v_0 t+y_0$. Let the firing point be the origin, so $y_0=0$. To complete the equation above, you must have an initial speed. 

To find the initial speed, apply the kinematic formula $y=-\frac 12 gt^{2}+v_0 t+y_0$ between the origin and striking point. \begin{gather*}y-y_0=-\frac 12 gt^{2}+v_0 t \\\\ -90-0=-\frac 12 (9.8)(10)^{2}+v_0(10) \\\\ \Rightarrow \boxed{v_0=40\,\rm m/s}\end{gather*} Notice that the striking point is $-90\,{\rm m}$ below the origin that's why the negative is entered for the vertical displacement $y$.  

Having the initial velocity, substitute it into the above equation again but with time $t=2\,{\rm s}$ to find the position of the bullet after $2\,{\rm s}$ with respect to the firing point \begin{align*}y-y_0&=-\frac 12 gt^{2}+v_0 t\\\\ &=-\frac 12\,(9.8)(2)^{2}+(40)(2) \\\\ \Rightarrow y&=\boxed{60.4\,\rm m} \end{align*}

Problem (16): A ball is thrown vertically upward with an initial velocity of $18\,\rm m/s$. How many seconds after throwing, the ball's speed is $9\,\rm m/s$ downward?

Solution : In all kinematic equations, $x,y,v,v_0$, and $a$ are vectors, so their signs matter. A speed of $9\,{\rm m/s}$ downward means a velocity of $-9\,{\rm m/s}$. Downward or upward indicates the direction of velocity. 

Take up as the positive $y$ direction. Now use the equation $v=v_0-gt$ to find the velocity at any later time.\begin{gather*}v=v_0-gt\\-9=+18-(10)t\\ \Rightarrow \quad \boxed{t=2.7\,\rm s}\end{gather*}

Problem (17): An object is thrown vertically upward in the air from a $100\,{\rm m}$ height with an initial velocity of $v_0$. After $5\,{\rm s}$, it reaches the ground. Determine the magnitude and direction of the initial velocity.

Solution : Let the origin be the throwing point (so $y_0=0$) and the upward direction positive. Substitute the given total time into the vertical displacement kinematic equation $y-y_0=-\frac 12 gt^2+v_0t$ with $y-y_0=-100\,{\rm m}$ (since the impact point is below the origin). \begin{gather*} y-y_0 =-\frac 12 gt^2+v_0t \\\\ -100=-\frac 12\,(10)(5)^2+v_0 (5) \\\\ \Rightarrow \boxed{v_0=-5\,\rm m/s}\end{gather*} the minus sign indicates the initial velocity is downward with the magnitude (speed) of $5\,\rm m/s$.

Problem (18): From a height of $15\,{\rm m}$, a ball is kicked vertically up into the air with an initial speed of $v_0$. It reaches the highest point of its path with an elevation of $20\,{\rm m}$ from the surface. Find the initial velocity $v_0$. 

Solution : The highest point is $5\,{\rm m}$ above the kicking point. Apply the time-independent kinematic equation below to find the initial velocity \begin{gather*}v^2-v_0^2=-2g(y-y_0) \\\\ 0-v_0^2=-2(10)(5) \\\\ \Rightarrow v_0=\pm 10\,{\rm m/s}\end{gather*} Because the ball kicked upward so we must choose the plus sign, i.e. $v_0=+10\,{\rm m/s}$. In the above, we used the fact that in all freely falling problems, at the highest point (apex) the velocity is zero ($v=0$).

Recall that projectiles are a particular type of freely falling motion with a launch angle of $\theta=90$ with its own formulas.

Problem (19): From the bottom of a $25\,{\rm m}$-depth well, a stone is thrown vertically upward with an initial speed of $30\,{\rm m/s}$.  (a) How high does the stone rise out of the well?  (b) Before the stone returns into the well, how many seconds are outside the well?

Solution :  (a) Let the bottom of the well be the origin, so $y_0=0$. First, we find how much distance the ball rises. Recall that the highest point is where $v=0$, so we have \begin{gather*}v^2-v_0^2=-2g(y-y_0) \\\\ 0-(30)^2=-2(10)(y-0) \\\\ \Rightarrow \boxed{y=45\,\rm m}\end{gather*} Of this height, $25\,{\rm m}$ is for the well's depth, so the stone is $20\,{\rm m}$ outside of the well. 

(b) We want to examine the duration between exiting and reentering the stone into the well. During this time interval, the ball returns to its initial position, so its displacement vector is zero, i.e., $\Delta y=y-y_0=0$. If we want to use the equation, $y-y_0=-\frac 12 gt^2+v_0t$, the speed of the stone is required exactly when it leaves the well. 

The speed at the bottom of the well is known. Thus, apply the equation $v^2-v_0^2=-2g\Delta y$ to find the speed just before it leaves the well. \begin{gather*}v^2-v_0^2=-2g\Delta y \\\\ v^2-(30)^{2}=-2(10)(25) \\\\ \Rightarrow v=+20\,{\rm m/s}\end{gather*} This speed can be used as the initial speed for the part where the stone is outside the well. Hence, the total time the stone is out of the well is obtained as below \begin{gather*} \Delta y=-\frac 12 gt^{2}+v_0 t\\ 0=-\frac 12 (10)t^{2}+(20)(2) \end{gather*} Solving for $t$, one can obtain the required time as $t=4\,{\rm s}$.

Problem (20): From the top of a $20-{\rm m}$-high tower, a small ball is thrown vertically upward. If $4\,{\rm s}$ after throwing, it hit the ground. How many seconds before striking the surface does the ball again meet the original throwing point? (Air resistance is neglected and $g=10\,{\rm m/s^2}$). 

Solution : Let the origin be the throwing point. The ball strikes the ground $20\,\rm m$ below our chosen origin, so its total displacement between initial position $y_i=0$ and final position is $\Delta y=y_f-y_i=-20\,\rm m$. The total time in which the ball is in the air is also $4\,{\rm s}$. With these known values, one can find the initial velocity as \begin{gather*}\Delta y=-\frac 12 gt^{2}+v_0t \\\\ -25=-\frac 12 (10)(4)^{2}+v_0(4) \\\\ \Rightarrow \boxed{v_0=15\,\rm m/s} \end{gather*} When the ball returns to its initial position, its total displacement is zero, i.e., $\Delta y=0$ so we can use the following kinematic equation to find the total time it takes the ball to return to the starting point \begin{gather*}\Delta y=-\frac 12 gt^{2}+v_0t \\\\ 0=-\frac 12 (10)t^{2}+(15)t \end{gather*} Rearranging and solving for $t$, we get $t=3\,{\rm s}$.

Free fall problem from a height

Problem (21): A rock is thrown vertically upward in the air. It reaches the height of $40\,{\rm m}$ from the surface at times $t_1=2\,{\rm s}$ and $t_2$. Find $t_2$ and determine the greatest height reached by the rock (neglect air resistance and assume $g=10\,{\rm m/s^2}$).

Solution : Let the throwing point (surface of the ground) be the origin. Between our chosen origin and the point with known values $h=4\,{\rm m}$, $t=2\,{\rm s}$ one can write down the kinematic equation $\Delta y=-\frac 12 gt^{2}+v_0\,t$ to find the initial velocity as \begin{gather*} \Delta y=-\frac 12 gt^{2}+v_0t \\\\ 40=-\frac 12 (10)(2)^2 +v_0(2) \\\\ \Rightarrow v_0=30\,{\rm m/s}\end{gather*} Now we are going to find the times when the rock reaches the height $40\,{\rm m}$ (Recall that when an object is thrown upward, it passes through every point twice). Applying the same equation above, we get \begin{gather*} \Delta y=-\frac 12 gt^{2}+v_0t \\\\ 40=-\frac 12\,(10)t^2+30t \\\\ \Rightarrow \boxed{5t^2-30t+40=0} \end{gather*} Rearranging and solving for $t$ using quadratic equation formula, two times are obtained i.e. $t_1=2\,{\rm s}$ and $t_2=4\,{\rm s}$. Thus, again after $4\,\rm s$ the ball again is at the same height of $40\,\rm m$ from the surface. 

The greatest height is where the vertical velocity becomes zero so we have \begin{gather*}v^2-v_0^2 =-2g\Delta y \\\\ 0-(30)^2=-2(10)\Delta y\\\\ \Rightarrow \boxed{\Delta y=45\,\rm m}\end{gather*} Thus, the highest point where the rock can reach is located at $H=45\,{\rm m}$ above the ground.

Problem (22): A ball is launched with an initial velocity of $30\,{\rm m/s}$ straight upward. How long will it take the ball to reach $20\,{\rm m}$ below the highest point for the first time? (neglect air resistance and assume $g=10\,{\rm m/s^2}$).

Solution : Between the origin (surface level) and the highest point ($v=0$) apply the time-independent kinematic equation below to find the greatest height $H$ where the ball reaches.\begin{gather*}v^2-v_0^2=-2g\Delta y \\\\ 0-(30)^2=-2(10)H \\\\ \Rightarrow \boxed{H=45\,\rm m}\end{gather*} This is the maximum height that the ball can reach. The $20\,{\rm m}$ below this maximum height $H$ has a height of $h=45-20=25\,{\rm m}$. Now use the vertical displacement kinematic equation between the throwing point and the desired position to find the required time taken. \begin{gather*} \Delta y=-\frac 12 gt^2+v_0t \\\\ 25=-\frac 12(10)t^2+30(t)\end{gather*} Solving for $t$ (using quadratic formula), we get $t_1=1\,{\rm s}$ and $t_2=5\,{\rm s}$ one for up way and the second for down way.  

Problem (23): A stone is launched directly upward from the surface level with an initial velocity of $20\,{\rm m/s}$. How many seconds after launch is the stone's velocity $5\,{\rm m/s}$ downward?

Solution : Let the origin be at the surface level and take the positive direction up. Therefore, we have initial velocity $v_0=+20\,{\rm m/s}$ and final velocity $v=-5\,{\rm m/s}$. Use the velocity kinematic equation $v=v_0-gt$ to find the desired time as below \begin{gather*}v=v_0-gt \\-5=+20-10\times t \\ \Rightarrow \boxed{t=2.5\,\rm s}\end{gather*}

Problem (24): From a $25-{\rm m}$ building, a ball is thrown vertically upward at an initial velocity of $20\,{\rm m/s}$. How long will it take the ball to hit the ground?

Solution : Origin is considered to be at the throwing point, so $y_0=0$. Apply the position kinematic equation below to find the desired time \begin{gather*} y-y_0=-\frac 12 gt^2+v_0 t \\\\ -25=-\frac 12(10)t^2+20t \\\\ \Rightarrow 5t^2-20t-25=0 \end{gather*} Rearranging and converting it into the standard form of a quadratic equation $at^2+bt+c=0$, its solutions are obtained as \begin{align*}t_{1,2}&=\frac{-b\pm \sqrt{b^2-4\,ac}}{2a} \\\\ &=\frac{-(-4)\pm\sqrt{(-4)^2-4(1)(-5)}}{2(1)}\\\\ &=-1 \, \text{and} \, 5 \end{align*} Therefore, the time needed for the ball to hit the ground is $5\,{\rm s}$.

Problem (25): From the top of a building with a height of $60\,{\rm m}$, a rock is thrown directly upward at an initial velocity of $20\,{\rm m/s}$. What is the rock's velocity at the instant of hitting the ground? 

Solution : Apply the time-independent kinematic equation as \begin{gather*} v^2-v_0^2 =-2g(y-y_0) \\\\ v^2-(20)^2 =-2(10)(-60) \\\\ v^2 =1600\\\\ \Rightarrow \quad \boxed{v=\pm 40\,\rm m/s} \end{gather*} Therefore, the rock's velocity when it hit the ground is $v=-40\,{\rm m/s}$.

In this article, you learned how to solve free fall problems step-by-step using simple kinematic equations. 

Author : Dr. Ali Nemati Published : 8/11/2022  

© 2015 All rights reserved. by Physexams.com

free fall problem with solution

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1.1: Free Fall

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  • Page ID 91046

  • Russell Herman
  • University of North Carolina Wilmington

In this chapter we will study some common differential equations that appear in physics. We will begin with the simplest types of equations and standard techniques for solving them We will end this part of the discussion by returning to the problem of free fall with air resistance. We will then turn to the study of oscillations, which are modeled by second order differential equations.

Free fall example.

Let us begin with a simple example from introductory physics. Recall that free fall is the vertical motion of an object solely under the force of gravity. It has been experimentally determined that an object near the surface of the Earth falls at a constant acceleration in the absence of other forces, such as air resistance. This constant acceleration is denoted by \(-g\) , where \(g\) is called the acceleration due to gravity. The negative sign is an indication that we have chosen a coordinate system in which up is positive.

We are interested in determining the position, \(y(t)\) , of the falling body as a function of time. From the definition of free fall, we have

\[\ddot{y}(t)=-g \label{1.1}. \]

Note that we will occasionally use a dot to indicate time differentiation.

Differentiation with respect to time is often denoted by dots instead of primes.

This notation is standard in physics and we will begin to introduce you to this notation, though at times we might use the more familiar prime notation to indicate spatial differentiation, or general differentiation.

In Equation \(\PageIndex{1}\) we know \(g\) . It is a constant. Near the Earth’s surface it is about \(9.81 \mathrm{~m} / \mathrm{s}^{2}\) or \(32.2 \mathrm{ft} / \mathrm{s}^{2}\) . What we do not know is \(y(t)\) . This is our first differential equation. In fact it is natural to see differential equations appear in physics often through Newton’s Second Law, \(F=m a\) , as it plays an important role in classical physics. We will return to this point later.

So, how does one solve the differential equation in \(\PageIndex{1}\)? We do so by using what we know about calculus. It might be easier to see when we put in a particular number instead of \(g\) . You might still be getting used to the fact that some letters are used to represent constants. We will come back to the more general form after we see how to solve the differential equation.

\[\ddot{y}(t)=5. \nonumber \]

Recalling that the second derivative is just the derivative of a derivative, we can rewrite this equation as

\[\dfrac{d}{d t}\left(\dfrac{d y}{d t}\right)=5 \nonumber \]

This tells us that the derivative of \(d y / d t\) is 5 . Can you think of a function whose derivative is 5 ? (Do not forget that the independent variable is \(t\) .) Yes, the derivative of \(5 t\) with respect to \(t\) is 5 . Is this the only function whose derivative is 5 ? No! You can also differentiate \(5 t+1,5 t+\pi, 5 t-6\) , etc. In general, the derivative of \(5 t+C\) is 5, where \(C\) is an arbitrary integration constant.

So, Equation \(\PageIndex{2}\) can be reduced to

\[\dfrac{d y}{d t}=5 t+C \nonumber \]

Now we ask if you know a function whose derivative is \(5 t+C\) . Well, you might be able to do this one in your head, but we just need to recall the Fundamental Theorem of Calculus, which relates integrals and derivatives. Thus, we have

\[y(t)=\dfrac{5}{2} t^{2}+C t+D \nonumber \]

where \(D\) is a second integration constant.

Equation \(\PageIndex{5}\) gives the solution to the original differential equation. That means that when the solution is placed into the differential equation, both sides of the differential equation give the same expression. You can always check your answer to a differential equation by showing that your solution satisfies the equation. In this case we have

\(\ddot{y}(t)=\dfrac{d^{2}}{d t^{2}}\left(\dfrac{5}{2} t^{2}+C t+D\right)=\dfrac{d}{d t}(5 t+C)=5\)

Therefore, Equation \(\PageIndex{5}\) gives the general solution of the differential equation.

We also see that there are two arbitrary constants, \(C\) and \(D .\) Picking any values for these gives a whole family of solutions. As we will see, the equation \(\ddot{y}(t)=5\) is a linear second order ordinary differential equation. The general solution of such an equation always has two arbitrary constants.

Let’s return to the free fall problem. We solve it the same way. The only difference is that we can replace the constant 5 with the constant \(-g .\) So, we find that

\[\dfrac{d y}{d t}=-g t+C \nonumber \]

\[y(t)=-\dfrac{1}{2} g t^{2}+C t+D \nonumber \]

Once you get down the process, it only takes a line or two to solve.

There seems to be a problem. Imagine dropping a ball that then undergoes free fall. We just determined that there are an infinite number of solutions for the position of the ball at any time! Well, that is not possible. Experience tells us that if you drop a ball you expect it to behave the same way every time. Or does it? Actually, you could drop the ball from anywhere. You could also toss it up or throw it down. So, there are many ways you can release the ball before it is in free fall producing many different paths, \(y(t)\) . That is where the constants come in. They have physical meanings.

If you set \(t=0\) in the equation, then you have that \(y(0)=D .\) Thus, \(D\) gives the initial position of the ball. Typically, we denote initial values with a subscript. So, we will write \(y(0)=y_{0}\) . Thus, \(D=y_{0}\) .

That leaves us to determine \(C\) . It appears at first in Equation \(\PageIndex{6}\). Recall that \(\dfrac{d y}{d t}\) , the derivative of the position, is the vertical velocity, \(v(t)\) . It is positive when the ball moves upward. We will denote the initial velocity \(v(0)=v_{0} .\) Inserting \(t=0\) in Equation \(\PageIndex{6}\), we find that \(\dot{y}(0)=C\) . This implies that \(C=v(0)=v_{0}\) .

Putting this all together, we have the physical form of the solution for free fall as

\[y(t)=-\dfrac{1}{2} g t^{2}+v_{0} t+y_{0} \nonumber \]

Doesn’t this equation look familiar? Now we see that the infinite family of solutions consists of free fall resulting from initially dropping a ball at position \(y_{0}\) with initial velocity \(v_{0}\) . The conditions \(y(0)=y_{0}\) and \(\dot{y}(0)=v_{0}\) are called the initial conditions. A solution of a differential equation satisfying a set of initial conditions is often called a particular solution. Specifying the initial conditions results in a unique solution.

So, we have solved the free fall equation. Along the way we have begun to see some of the features that will appear in the solutions of other problems that are modeled with differential equation. Throughout the book we will see several applications of differential equations. We will extend our analysis to higher dimensions, in which we case will be faced with socalled partial differential equations, which involve the partial derivatives of functions of more that one variable.

But are we done with free fall? Not at all! We can relax some of the conditions that we have imposed. We can add air resistance. We will visit this problem later in this chapter after introducing some more techniques. We can also provide a horizontal component of motion, leading to projectile motion.

clipboard_ed7fc5030cd23ea32a777b9c19f8b980b.png

Finally, we should also note that free fall at constant \(g\) only takes place near the surface of the Earth. What if a tile falls off the shuttle far from the surface of the Earth? It will also fall towards the Earth. Actually, the tile also has a velocity component in the direction of the motion of the shuttle. So, it would not necessarily take radial path downwards. For now, let’s ignore that component. To look at this problem in more detail, we need to go to the origins of the acceleration due to gravity. This comes out of Newton’s Law of Gravitation. Consider a mass \(m\) at some distance \(h(t)\) from the surface of the (spherical) Earth. Letting \(M\) and \(R\) be the Earth’s mass and radius, respectively, Newton’s Law of Gravitation states that

\[ \begin{aligned} m a &=F \\ m \dfrac{d^{2} h(t)}{d t^{2}} &=-G \dfrac{m M}{(R+h(t))^{2}} \end{aligned} \nonumber \]

Here \(G=6.6730 \times 10^{-11} \mathrm{~m}^{3} \mathrm{~kg}^{-1} \mathrm{~s}^{-2}\) is the Universal Gravitational Constant, \(M=5.9736 \times 10^{24} \mathrm{~kg}\) and \(R=6371 \mathrm{~km}\) are the Earth’s mass and mean radius, respectively. For \(h<<R, G M / R^{2} \approx g\) .

Thus, we arrive at a differential equation

\[\dfrac{d^{2} h(t)}{d t^{2}}=-\dfrac{G M}{(R+h(t))^{2}} . \nonumber \]

This equation is not as easy to solve. We will leave it as a homework exercise for the reader.

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Mechanics (Essentials) - Class 11th

Course: mechanics (essentials) - class 11th   >   unit 4.

  • Sign of gravity in free fall
  • Free fall 1 body - solved example

Free fall - 2 body solved numerical

  • Freefall: graphs and conceptual questions
  • Free fall - total time up & down solved example
  • Solving freefall problems using kinematic formulas
  • Worked example: Free fall, object thrown up from a building
  • Advanced: Freefall problems

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Video transcript

Free Fall Calculator

Table of contents

This free fall calculator is a tool for finding the velocity of a falling object along with the distance it travels. Thanks to this tool, you can apply the free fall equation for any object, be it an apple you drop or a person skydiving.

Read on to learn the free fall definition and discover the most daring examples, including the highest free fall in history (spoiler alert: it broke the sound barrier )! We'll also explain what free fall acceleration is and why we assume it's constant.

Check out the projectile motion calculator , which describes a case of free fall combined with horizontal motion.

Prefer watching rather than reading? Check out our deep dive into the concept of free fall here:

What is the free fall definition?

In free fall, an object moves under the influence of gravitational force only . The only acceleration is the acceleration of gravity g . No other force, including air resistance, is acting on such an object.

Interestingly, an object in free fall doesn't necessarily need to be falling (that is, moving downwards). For example, the Moon's motion satisfies all of the conditions listed above: there is no other force acting on it other than gravity (it's being pulled towards the Earth), and there is no air resistance, as there is no air in space.

Why doesn't the Moon crash into Earth, then? It's because Moon's speed is not directed towards Earth, but tangentially to its orbit. Since the Moon is moving along an elliptic orbit with a high enough velocity, its motion generates a centrifugal force, equal and opposite to the force of gravity.

You may want to check out our gravitational force calculator and see what an amazing force gravity is.

Free fall speed

From the definition of velocity, we can find the velocity of a falling object is:

  • v 0 v_0 v 0 ​ – Initial velocity (measured in m/s or ft/s);
  • t t t – Fall time (measured in seconds); and
  • g g g – Free fall acceleration (expressed in m/s² or ft/s²).

Without the effect of air resistance, each object in free fall would keep accelerating by 9.80665 m/s (approximately equal to 32.17405 ft/s ) every second. In reality, though, a falling object's velocity is constrained by a value called the terminal velocity .

What is the terminal velocity? As you have seen above, the free-fall acceleration is constant, which means that the gravitational force acting on an object is constant, too. However, the force of air resistance increases with increasing free fall speed. At some point, the two forces become equal in magnitude . According to Newton's first law, at that point, the falling body stops accelerating and moves at a constant speed. This speed is the terminal velocity.

In this free fall calculator, we neglect the influence of air resistance. If you want to consider it, head over to our free fall with air resistance calculator .

Free fall equation

If you want to calculate the distance traveled by a falling object, you need to write down the equation of motion. If the initial displacement and velocity are both equal to zero, it boils down to:

If the object is already traveling with an initial velocity, you have to take it into account, too:

You can immediately see that the object's distance traveled is proportional to the fall time squared. It means that with each second, the falling body travels a substantially larger distance than before.

Another interesting fact is that according to the free fall formula, the distance does not depend on the mass of the falling object . If you drop a feather and a brick, they will hit the ground at the same time… Or at least that's what science says! If you try to perform an experiment, you'll notice that, in reality, the brick falls to the ground first. Why does that happen? Again, because of air resistance. If you dropped the two items in a vacuum, they would both hit the ground at the same instant!

How to use the free fall formula: an example

Still not sure how our free fall calculator works? Don't worry – we prepared a simple example to walk you through it.

Determine the gravitational acceleration. On Earth, this value is equal to 9.80665 m/s² on average (which is also the default value set in the free fall calculator).

Decide whether the object has an initial velocity. We will assume v₀ = 0 m/s .

Choose how long the object is falling. In this example, we will use the time of 8 seconds .

Calculate the final free fall speed (just before hitting the ground) with the formula:

v = v₀ + gt = 0 + 9.80665 × 8 = 78.45 m/s .

Find the free fall distance using the equation:

s = (1/2)gt² = 0.5 × 9.80665 × 8² = 313.8 m .

If you know the height from which the object is falling, but don't know the time of fall, you can use this calculator to find it, too!

Highest free fall in history

You might already have learned the free fall equation, but it's one thing to understand the theory and a completely different one to experience it. There are many ways to experience the thrill of a free fall – you could, for example, jump with a parachute or try bungee jumping!

Technically, such a jump doesn't fulfill all the requirements of a free fall – there is substantial air resistance involved. In fact, a real free fall is only possible in a vacuum. Nevertheless, this is as close to the actual experience as you can get on Earth 😉

One of the most extreme examples of an almost-scientifically correct free fall is the jump of Dr. Alan Eustace, Google's VP of Knowledge, in 2014. Eustace jumped from a heart-stopping height of 135,908 feet (41,425 m) , thus setting a new record for a parachute jump.

Surprisingly, Eustace declined Google's help in the jump and funded the project himself. It was not an easy endeavor because such a leap required him to go up in a special balloon and wear a custom-designed spacesuit that protected him from sudden shifts in temperature (after all, he was jumping from the edge of space). The fall itself took 15 minutes, and the maximum speed exceeded 800 miles per hour – far over the sound barrier!

What is free fall speed?

Why is the weight of a free falling body zero.

It is not. An object in free fall will still have a weight , governed by the equation W = mg , where W is the object’s weight, m is the object’s mass, and g is the acceleration due to gravity. Weight, however, does not affect an object's free-falling speed . Two identically shaped objects weighing a different amount will hit the ground at the same time.

What is the difference between free fall and weightlessness?

Free fall is when an object is falling , only being affected by the force of gravity, while weightlessness is when an object has no weight due to there being no effect from gravity (it still has mass). Weightlessness can be achieved either in space or if an equal force can be applied in the opposite direction of gravity.

How do you find free fall acceleration of a planet?

To find the free fall acceleration of a planet:

Estimate the total mass of the planet in kilograms.

Find the radius of the planet , from its center to its surface, in meters.

Divide the total mass by the radius squared.

Multiply the result by the universal Gravitational constant :

6.67×10 -11 N·m 2 ·kg -2

The result is the gravitational acceleration of the planet, which is also its free-fall acceleration.

Wondering how many helium balloons it would take to lift you up in the air? Try this helium balloons calculator! 🎈

Check out the impact meat has on the environment and your health.

Do you feel like you could be doing something more productive or educational while on a bus? Or while cleaning the house? Well, why don't you dive into the rich world of podcasts! With this podcast calculator, we'll work out just how many great interviews or fascinating stories you can go through by reclaiming your 'dead time'!

Gravitational acceleration (g)

Initial velocity (v₀)

Time of fall (t)

Velocity (v)

Velocity at time t.

Read our research on: Immigration & Migration | Podcasts | Election 2024

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How americans view the situation at the u.s.-mexico border, its causes and consequences, 80% say the u.s. government is doing a bad job handling the migrant influx.

free fall problem with solution

Pew Research Center conducted this study to understand the public’s views about the large number of migrants seeking to enter the U.S. at the border with Mexico. For this analysis, we surveyed 5,140 adults from Jan. 16-21, 2024. Everyone who took part in this survey is a member of the Center’s American Trends Panel (ATP), an online survey panel that is recruited through national, random sampling of residential addresses. This way nearly all U.S. adults have a chance of selection. The survey is weighted to be representative of the U.S. adult population by gender, race, ethnicity, partisan affiliation, education and other categories. Read more about the ATP’s methodology .

Here are the questions used for the report and its methodology .

The growing number of migrants seeking entry into the United States at its border with Mexico has strained government resources, divided Congress and emerged as a contentious issue in the 2024 presidential campaign .

Chart shows Why do Americans think there is an influx of migrants to the United States?

Americans overwhelmingly fault the government for how it has handled the migrant situation. Beyond that, however, there are deep differences – over why the migrants are coming to the U.S., proposals for addressing the situation, and even whether it should be described as a “crisis.”

Factors behind the migrant influx

Economic factors – either poor conditions in migrants’ home countries or better economic opportunities in the United States – are widely viewed as major reasons for the migrant influx.

About seven-in-ten Americans (71%), including majorities in both parties, cite better economic opportunities in the U.S. as a major reason.

There are wider partisan differences over other factors.

About two-thirds of Americans (65%) say violence in migrants’ home countries is a major reason for why a large number of immigrants have come to the border.

Democrats and Democratic-leaning independents are 30 percentage points more likely than Republicans and Republican leaners to cite this as a major reason (79% vs. 49%).

By contrast, 76% of Republicans say the belief that U.S. immigration policies will make it easy to stay in the country once they arrive is a major factor. About half as many Democrats (39%) say the same.

For more on Americans’ views of these and other reasons, visit Chapter 2.

How serious is the situation at the border?

A sizable majority of Americans (78%) say the large number of migrants seeking to enter this country at the U.S.-Mexico border is eithera crisis (45%) or a major problem (32%), according to the Pew Research Center survey, conducted Jan. 16-21, 2024, among 5,140 adults.

Related: Migrant encounters at the U.S.-Mexico border hit a record high at the end of 2023 .

Chart shows Border situation viewed as a ‘crisis’ by most Republicans; Democrats are more likely to call it a ‘problem’

  • Republicans are much more likely than Democrats to describe the situation as a “crisis”: 70% of Republicans say this, compared with just 22% of Democrats.
  • Democrats mostly view the situation as a major problem (44%) or minor problem (26%) for the U.S. Very few Democrats (7%) say it is not a problem.

In an open-ended question , respondents voice their concerns about the migrant influx. They point to numerous issues, including worries about how the migrants are cared for and general problems with the immigration system.

Yet two concerns come up most frequently:

  • 22% point to the economic burdens associated with the migrant influx, including the strains migrants place on social services and other government resources.
  • 22% also cite security concerns. Many of these responses focus on crime (10%), terrorism (10%) and drugs (3%).

When asked specifically about the impact of the migrant influx on crime in the United States, a majority of Americans (57%) say the large number of migrants seeking to enter the country leads to more crime. Fewer (39%) say this does not have much of an impact on crime in this country.

Republicans (85%) overwhelmingly say the migrant surge leads to increased crime in the U.S. A far smaller share of Democrats (31%) say the same; 63% of Democrats instead say it does not have much of an impact.

Government widely criticized for its handling of migrant influx

For the past several years, the federal government has gotten low ratings for its handling of the situation at the U.S.-Mexico border. (Note: The wording of this question has been modified modestly to reflect circumstances at the time).

Chart shows Only about a quarter of Democrats and even fewer Republicans say the government has done a good job dealing with large number of migrants at the border

However, the current ratings are extraordinarily low.

Just 18% say the U.S. government is doing a good job dealing with the large number of migrants at the border, while 80% say it is doing a bad job, including 45% who say it is doing a very bad job.

  • Republicans’ views are overwhelmingly negative (89% say it’s doing a bad job), as they have been since Joe Biden became president.
  • 73% of Democrats also give the government negative ratings, the highest share recorded during Biden’s presidency.

For more on Americans’ evaluations of the situation, visit Chapter 1 .

Which policies could improve the border situation?

There is no single policy proposal, among the nine included on the survey, that majorities of both Republicans and Democrats say would improve the situation at the U.S.-Mexico border. There are areas of relative agreement, however.

A 60% majority of Americans say that increasing the number of immigration judges and staff in order to make decisions on asylum more quickly would make the situation better. Only 11% say it would make things worse, while 14% think it would not make much difference.

Nearly as many (56%) say creating more opportunities for people to legally immigrate to the U.S. would make the situation better.

Chart shows Most Democrats and nearly half of Republicans say boosting resources for quicker decisions on asylum cases would improve situation at Mexico border

Majorities of Democrats say each of these proposals would make the border situation better.

Republicans are less positive than are Democrats; still, about 40% or more of Republicans say each would improve the situation, while far fewer say they would make things worse.

Opinions on other proposals are more polarized. For example, a 56% majority of Democrats say that adding resources to provide safe and sanitary conditions for migrants arriving in the U.S. would be a positive step forward.

Republicans not only are far less likely than Democrats to view this proposal positively, but far more say it would make the situation worse (43%) than better (17%).

Chart shows Wide partisan gaps in views of expanding border wall, providing ‘safe and sanitary conditions’ for migrants

Building or expanding a wall along the U.S.-Mexico border was among the most divisive policies of Donald Trump’s presidency. In 2019, 82% of Republicans favored expanding the border wall , compared with just 6% of Democrats.

Today, 72% of Republicans say substantially expanding the wall along the U.S. border with Mexico would make the situation better. Just 15% of Democrats concur, with most saying either it would not make much of a difference (47%) or it would make things worse (24%).

For more on Americans’ reactions to policy proposals, visit Chapter 3 .

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Table of contents, fast facts on how greeks see migrants as greece-turkey border crisis deepens, americans’ immigration policy priorities: divisions between – and within – the two parties, from the archives: in ’60s, americans gave thumbs-up to immigration law that changed the nation, around the world, more say immigrants are a strength than a burden, latinos have become less likely to say there are too many immigrants in u.s., most popular.

About Pew Research Center Pew Research Center is a nonpartisan fact tank that informs the public about the issues, attitudes and trends shaping the world. It conducts public opinion polling, demographic research, media content analysis and other empirical social science research. Pew Research Center does not take policy positions. It is a subsidiary of The Pew Charitable Trusts .

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Why the Most Educated People in America Fall for Anti-Semitic Lies

At Harvard and elsewhere, an old falsehood is capturing new minds.

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B y now, December’s congressional hearing about anti-Semitism at universities, during which the presidents of Harvard, the University of Pennsylvania, and MIT all claimed that calls for the genocide of Jews would violate their university’s policies only “depending on the context,” is already a well-worn meme. Surely there is nothing left to say about this higher-education train wreck, after the fallout brought down two of those university presidents and spawned a thousand op-eds—except that all of the punditry about diversity and free speech and criticism of Israel has extravagantly missed the point.

The problem was not that Jewish students on American university campuses didn’t want free speech, or that they didn’t want to hear criticism of Israel. Instead, they didn’t want people vandalizing Jewish student organizations’ buildings, or breaking or urinating on the buildings’ windows. They didn’t want people tearing their mezuzahs down from their dorm-room doors. They didn’t want their college instructors spouting anti-Semitic lies and humiliating them in class. They didn’t want their posters defaced with Hitler caricatures, or their dorm windows plastered with Fuck Jews . They didn’t want people punching them in the face, or beating them with a stick, or threatening them with death for being Jewish. At world-class American colleges and universities, all of this happened and more.

I was not merely an observer of this spectacle. I’d been serving on now–former Harvard President Claudine Gay’s anti-Semitism advisory committee, convened after the October 7 Hamas massacre in Israel and amid student responses to it. I was asked to participate because I am a Harvard alumna who wrote a book about anti-Semitism called People Love Dead Jews . As soon as my participation became public, I was inundated with messages from Jewish students seeking help. They approached me with their stories after having already tried many other avenues—bewildered not only by what they’d experienced, but also by how many people dismissed or denied those experiences.

In Congress, all three university presidents offered some version of the platitudes that “Hatred comes from ignorance” and “Education is the answer.” But if hatred comes from ignorance, why were America’s best universities full of this very specific ignorance? And why were so many people trying to justify it, explain it away, or even deny it? Our era’s 10-second news cycle is no match for these questions, because the answers are deep and ancient, buried beneath the oldest of assumptions about what we think we know.

Read: What Claudine Gay got right and the International Court of Justice got wrong

The through line of anti-Semitism for thousands of years has been the denial of truth and the promotion of lies. These lies range in scope from conspiracy theories to Holocaust denial to the blood libel to the currently popular claims that Zionism is racism, that Jews are settler colonialists, and that Jewish civilization isn’t indigenous to the land of Israel. These lies are all part of the foundational big lie: that anti-Semitism itself is a righteous act of resistance against evil, because Jews are collectively evil and have no right to exist. Today, the big lie is winning.

I n 2013, David Nirenberg published an astonishing book titled Anti-Judais m . Nirenberg’s argument, rigorously laid out in nearly 500 pages of dense scholarship and more than 100 pages of footnotes, is that Western cultures—including ancient civilizations, Christianity, Islam (which Nirenberg considers Western in its relationship with Judaism), and post-religious societies—have often defined themselves through their opposition to what they consider “Judaism.” This has little to do with actual Judaism, and a lot to do with whatever evil these non-Jewish cultures aspire to overcome.

Nirenberg is a diligent historian who resists generalizations and avoids connecting the past to contemporary events. But when one reads through his carefully assembled record of 23 centuries’ worth of intellectual leaders articulating their societies’ ideals by loudly rejecting whatever they consider “Jewish,” this deep neural groove in Western thought becomes difficult to dismiss, its patterns unmistakable. If piety was a given society’s ideal, Jews were impious blasphemers; if secularism was the ideal, Jews were backward pietists. If capitalism was evil, Jews were capitalists; if communism was evil, Jews were communists. If nationalism was glorified, Jews were rootless cosmopolitans; if nationalism was vilified, Jews were chauvinistic nationalists. “Anti-Judaism” thus becomes a righteous fight to promote justice.

This dynamic forces Jews into the defensive mode of constantly proving they are not evil, and even simply that they have a right to exist. Around 38 C.E., after rioters in Alexandria destroyed hundreds of Jewish homes and burned Jews alive, the Jewish Alexandrian intellectual Philo and the non-Jewish Alexandrian intellectual Apion both sailed to Rome for a “debate” before Emperor Caligula about whether Jews deserved citizenship. Apion believed that Jews held an annual ritual in which they kidnapped a non-Jew, fattened him up, and ate him. Caligula delayed Philo’s rebuttal for five months, and then listened to him only while consulting with designers on palace decor. Alexandrian Jews lost their citizenship rights, though it took until 66 C.E. for 50,000 more of them to be slaughtered.

In medieval Europe, Jews were forced into disputations with Christian priests that placed Jewish texts and traditions on public trial, resulting in Jewish books being burned and Jewish disputants exiled. Later legal trials expanded on this concept, requiring Jews to defend themselves against the absurd charge known as the blood libel, in which Jews are accused of murdering and consuming non-Jewish children—a claim that has echoes in current lies about Israelis harvesting Palestinians’ organs.

The absurdity of these charges is less remarkable than the high intellectual profiles of those making them: people like Apion, a scholar of Homer and Egyptian history, as well as Christian and Muslim scholars who were among the best-read people of their time. Similarly absurd claims of Jewish perfidy were later endorsed by civilizational luminaries such as Martin Luther and Voltaire. “Anti-Judaism,” Nirenberg argues, “should not be understood as some archaic or irrational closet in the vast edifices of Western thought. It was rather one of the basic tools with which that edifice was constructed.”

protest at Harvard University

I’ve been thinking about Nirenberg’s thesis in the months since the October 7 massacre in Israel, during which Hamas, an openly genocidal organization whose stated goal is the murder of Jews, lived up to its mission statement by torturing, raping, and murdering more than 1,200 people in southern Israel and taking more than 200 captives, including babies, children, and the elderly. Shortly after the attacks, a Cornell professor publicly proclaimed the barbarity “exhilarating” and “energizing,” while a Columbia professor called it “awesome” and an “achievement.” Comparable praise percolated through America’s top universities, coming from students and faculty alike. On campuses around the country, students began gathering regularly to chant “There is only one solution: intifada revolution!”—a reference to a suicide-bombing campaign in Israel a generation ago that maimed and murdered well over 1,000 Jews. (If there is only one solution, perhaps one could call it the Final Solution.)

Students took these rallies inside libraries and other campus buildings. They vandalized university property with such slogans as “Zionism = Genocide,” “New Intifada,” and “From the river to the sea, Palestine will be free”—referring to a geographic area that encompasses the entirety of the state of Israel, where half the world’s Jews live. (At Harvard, some students opted for chanting an Arabic version: “From water to water, Palestine is Arab.”) On some campuses, the exhilaration escalated into death threats and physical assaults against Jewish students. When a Jewish Tulane University student tried to stop an anti-Israel protester near campus from burning an Israeli flag, protesters attacked him and other Jewish students, breaking one student’s nose.

It wasn’t just universities. Crowds cheering for “intifada” gathered in cities around the country, shutting down and disrupting train stations and airport access roads. Lest their support for Hamas be mistaken for support for Palestinians in general, or for peace, U.S. rally organizers named their efforts “floods” (“Flood Seattle for Palestine,” “Flood Manhattan for Gaza”) after “Operation Al Aqsa Flood,” Hamas’s name for its October 7 butchery. The enthusiasm was hard to contain. Some people tore down or vandalized posters of Israeli hostages. Others targeted synagogues and Jewish-owned businesses, spray-painting them with swastikas and slogans like “Israel’s only religion is capitalism.” In New York City, a Jewish teacher’s online photo holding a sign that said I Stand With Israel was enough to prompt a schoolwide protest that devolved into a riot during which students destroyed school property; the teacher had to be moved to another part of the building to avoid the teenage mob screaming “Free Palestine!” In Los Angeles, a man invaded a Jewish family’s home before dawn with a knife, breaking into the parents’ bedroom while their four children slept, screaming “Kill Jewish people.” When police arrested him, he shouted, “Free Palestine!”

Criticism of Israel is not anti-Semitic : Jews are now required to recite this humiliatingly obvious sentence, over and over, as the price of admission to public discourse about their own demonization, in “debates” with people who are often unable to name the relevant river or sea. The many legitimate concerns about Israel’s policies toward Palestinians, and the many legitimate concerns about Israel’s current war in Gaza, cannot explain these eliminationist chants and slogans, the glee with which they are delivered, the lawlessness that has accompanied them, or the open assaults on Jews. The timing alone laid the game bare: This mass exhilaration first emerged not in response to Israel’s war to take down Hamas and rescue its kidnapped citizens, but exactly in response to, and explicitly in support of, the most lethal and sadistic barbarity against Jews since the Holocaust, complete with rape and decapitation and the abduction of infants, committed by a regime that aims to eviscerate not only Jews, but also all hopes of Palestinian flourishing, coexistence, or peace.

Read: When anti-Zionism is anti-Semitic

But there are nuances to sadistic barbarity against Jews, we are told, and sometimes gang-raping Jewish women is actually a movement for human rights. It hardly seems fair to call people anti-Semitic if they want only half of the world’s Jews to die. The phrase “Globalize the Intifada,” currently chanted at universities across America, perhaps widens the net a tiny bit—but really, who can say? Even the phrase “Gas the Jews,” chanted at a rally organized by NYU students and faculty, is so very ambiguous. How dare those whiny Jews presume to know what’s in other people’s hearts?

Besides, American Jews had nothing to whine about: Had any of them actually died in the United States from all this exhilaration? That question was answered in November, when a Jewish man died in California after an anti-Israel protester allegedly clubbed him over the head with a bullhorn, the kind used to chant entirely non-anti-Semitic slogans—and of course that question had already been answered repeatedly with other anti-Semitic murders in recent years, some more publicized than others. (One murder even happened on campus: In 2022, an expelled University of Arizona student who repeatedly ranted about Jews and Zionists shot and killed his professor—who wasn’t Jewish, though the student thought he was.) But now the goalposts move again: Those actual murders, along with many other physical attacks against American Jews, are all just one-offs, lone wolves, mental-illness cases, entirely unrelated to the anti-Semitic rhetoric swirling through American life.

It remains unclear why anti-Semitism should matter only when it is lethal, or if so, how many unambiguously anti-Semitic murders would be necessary for anti-Semitism to be happening outside whiny Jews’ heads. A realistic estimate might be 6 million. Even then, Jews have had to spend the past 80 years collecting documentation to prove it.

O ne confounding fact in this onslaught of the world’s oldest hatred is that American society should have been ready to handle it. Many public and private institutions have invested enormously in recent years in attempts to defang bigotry; ours is an era in which even sneaker companies feel obliged to publicly denounce hate. But diversity, equity, and inclusion initiatives have proved to be no match for anti-Semitism, for a clear reason: the durable idea of anti-Semitism as justice.

DEI efforts are designed to combat the effects of social prejudice by insisting on equity: Some people in our society have too much power and too much privilege, and are overrepresented, so justice requires leveling the playing field. But anti-Semitism isn’t primarily a social prejudice. It is a conspiracy theory: the big lie that Jews are supervillains manipulating others. The righteous fight for justice therefore does not require protecting Jews as a vulnerable minority. Instead it requires taking Jews down.

This idea is tacitly endorsed by Jews’ bizarre exclusion from discussion in many DEI trainings and even policies, despite their high ranking in American hate-crime statistics. The premise, for instance, that Jews don’t experience bigotry because they are “white,” itself a fraught idea, would suggest that white LGBTQ people don’t experience bigotry either—a premise that no DEI policy would endorse (not to mention the fact that many Jews are not white). The contention that Jews are immune to bigotry because they are “rich,” an idea even more fraught and also often false (about 20 percent of Jews in New York City, for instance, live in poverty or near-poverty), is equally nonsensical. No one claims that gay men or Indian Americans never experience bigotry because of those groups’ statistically higher incomes. The idea that money erases bigotry apparently applies only to Jews. Again and again, the ostensible reasons for not addressing anti-Semitism in DEI initiatives quickly reveal themselves to be founded on ancient, rarely examined assumptions about Jews as invulnerable villains.

The sordid history of the concept of anti-Zionism vividly illustrates this dynamic—and is particularly relevant for its success in scrambling the radar of well-meaning people. Jewish civilization has been centered for thousands of years, in ways large and small, on its homeland in Israel, where Jews have had a continuous presence since ancient times. The modern political idea of Zionism as Jewish self-determination in this homeland emerged in the late 19th and early 20th centuries amid many other anticolonial movements around the world, as global power dynamics shifted from empires (Habsburg, Russian, Ottoman, British, French, Japanese) toward nation-states. The large and often violent population upheavals following Israel’s creation, including the displacement of most Arabs from what became Israel and the displacement of nearly all Jews from what became Arab states, paralleled similar population upheavals around the world as new states emerged from receding empires. In this, Zionism was typical.

But anti-Zionism as an explicit political concept has a history quite independent of the actions of Jews. In 1918, 30 years before the establishment of the state of Israel, Bolsheviks established Jewish sections of the Communist Party, which they insisted be anti-Zionist. The problem, Bolsheviks argued, was that Jewish particularism (in this case, Zionism) was the obstacle to the righteous universal mission of uniting humanity under communism—just as Christians once saw Jewish particularism as the obstacle to the righteous universal mission of uniting humanity under Christ. The righteousness of this mission was, as usual, the key: The claim that “anti-Zionism” was unrelated to anti-Semitism, repeated ad nauseam in Soviet propaganda for decades, was essential to the Communist Party’s self-branding as humanity’s liberators. It was also a bald-faced lie.

Bolsheviks quickly demonstrated their supposed lack of anti-Semitism by shutting down every “Zionist” institution under their control, a category that ranged from synagogues to sports clubs; appropriating their assets; taking over their buildings, sometimes physically destroying offices; and arresting and ultimately “purging” Jewish leaders, including those who had endorsed the party line and persecuted their fellow Jews for their “Zionism.” Thousands of Jews were persecuted, imprisoned, tortured, or murdered.

Later, the U.S.S.R. exported this messaging to its client states in the developing world and ultimately to social-justice-minded circles in the United States. A thick paper trail shows how the KGB adapted its propaganda by explicitly rebranding Zionism as “racism” and “colonialism,” beginning half a century ago, when those terms gained currency as potent smears—even though Jews are racially diverse and Zionism is one of the world’s premier examples of an indigenous people reclaiming independence. Facts were irrelevant: Soviets labeled Jews as racist colonialist oppressors, just as Nazis had labeled Jews as both capitalist and Communist oppressors, and just as Christians and Muslims had labeled Jews as God-killers and Prophet-defilers. Jews were whatever a given society regarded as evil. To borrow the language of DEI, the big lie is systemic.

Even naming it—that is, calling out bigotry against Jews—can be classed as yet another sign of assumed evil intent, of Jews attacking beloved principles of justice for all. In an April 2023 lecture , David Nirenberg, the historian, presented the example of an activist with a large following whose boundary-pushing rhetoric met with accusations of anti-Semitism. The activist pointed out, as Nirenberg put it, that anti-Semitism “was merely an accusation that Jews used to silence criticism and squash free speech.” He brought libel lawsuits against newspapers that accused him of anti-Semitism, and won them. It is unfortunate for those making this argument today that this activist was named Adolf Hitler.

T wo weeks after the October 7 massacre, I wrote an op-ed for a national newspaper about the intergenerational fears many Jews were feeling, describing a few choice moments from several thousand years of anti-Semitic attacks. A friendly fact-checker followed up, asking me to prove that the Russian Civil War pogroms of 1918–21 involved gang rapes, and appending a judicious reportedly in front of a detail I’d included from the Farhud pogrom in Baghdad in 1941 about attackers taking Jewish women’s severed breasts as trophies. I dutifully provided additional sources, combing through sickening testimonies about mutilated Jewish girls in 1919 and 1941, while simultaneously avoiding videos of mutilated Jewish girls in 2023.

As I piled up evidence to prove that these things happened, I remembered an oral-history interview my sister once did with our grandfather to share with our family at his 97th-birthday party, in which he described his own grandparents’ decision to leave their town in Ukraine after an aunt was attacked during a pogrom. “They raided her, et cetera, et cetera,” my sister’s notes from the interview say. Et cetera, et cetera , I thought over and over, as I hunted down sources on gang rapes of Jewish women to submit to the fact-checker, my vision going blurry. At the time, I hadn’t wondered what those sanitized et cetera s meant.

The same week I spent emailing documentation to the fact-checker of pogroms long past, the newspaper, like many other news outlets, published a banner headline about Israelis bombing a hospital in Gaza and killing 500 people inside. This was quickly proven to be a lie told by Hamas—a lie similar to the medieval blood libel, about Jews deliberately targeting and murdering innocent non-Jewish babies—and a transparent psychological projection of the crimes that Hamas had actually committed in Israel, where Hamas terrorists had deliberately targeted and murdered hundreds of adults, children, and babies, and also repeatedly fired rockets at a hospital. Israel’s military has indeed killed many innocent people in Gaza during its war to destroy Hamas, and deserves the same scrutiny as any country for its conduct in war. But scrutiny is impossible when lies are substituted for facts. The newspaper later issued a regretful editorial note acknowledging its error. Unfortunately, Hamas’s lie had already inspired mass demonstrations around the world; rioters in Tunisia were so incensed by it that they burned a historic synagogue to the ground. I had been rightfully asked to prove that the Iraqi and Ukrainian pogroms happened. But the spokespeople for Hamas were taken at their word.

Shortly after the op-ed was published, I was invited to watch video footage of the October 7 attacks that the Israeli army had compiled from security cameras, online videos, and Hamas terrorists’ GoPro cameras. This grim footage was assembled specifically for the purpose of fighting back against denial. But even this horrifying and humiliating evidence, documented largely by the perpetrators themselves, apparently isn’t enough to prove that Jewish experiences are real. At a screening of the footage in Los Angeles, someone in the audience shouted , “Show the rapes!”

The attackers themselves provided footage of a woman’s naked, mutilated corpse and of a teenager with blood-soaked pants being dragged by her hair out of a truck. Since then, it has become clear that Hamas used rape and sexual torture systematically against Israeli women. Israeli first responders and forensic scientists have found corpses of women and girls with vaginal bleeding and broken pelvises. Teenage sisters were found murdered in their bedroom, one shot in the head with her pants pulled down, covered in semen; one woman was found with nails and other objects in her genitalia, while others were found to have been shot through their vaginas. Eyewitness testimony has included details about a woman who was passed among many men, murdered while one of them was still raping her; at one point, her severed breast was tossed in the air. It’s a detail familiar from the 1941 Baghdad pogrom, just as slicing a fetus out of a pregnant Jewish woman’s body is a tactic Hamas unknowingly replicated from the Khmelnytskyi pogroms of 1648 Ukraine. Et cetera, et cetera. But who would believe it? “Show the rapes!”

Graeme Wood: A record of pure, predatory sadism

I was invited to these screenings multiple times, but never went. I didn’t want to watch people being brutalized. Also, I didn’t want to watch people being brutalized while hearing someone behind me screaming, “Show the rapes!”

O n my travels around the country in recent months to discuss my work on Jews in non-Jewish societies, I met many Jewish college and high-school students who seem to have accepted the casual denigration of Jews as normal. They are growing up with it. In a Dallas suburb, teenagers told me, shrugging, about how their friends’ Jewish fraternities at Texas colleges have been “chalked.” I had to ask what “chalking” meant: anti-Semitic graffiti made by vandals who lacked spray paint. Synagogues are often chalked too. Another newly common verb among American Jews is swatting : fake bomb or active-shooter threats that force evacuations and instill fear. (The term is a reference to the SWAT teams that sometimes arrive at the scene, not knowing the threat is a hoax, and instill more fear.) These now happen so often at American Jewish institutions that they’re almost boring; nearly 200 were swatted during one December 2023 weekend alone. (When it happened at my own synagogue in November, forcing a girl’s bat-mitzvah service into a parking lot, the synagogue president warned congregants not to post any specific details about it online, in case people were tracking our evacuation procedures.)

Daniel Torday: What active-shooter trainings steal from synagogues

American Jews in recent years have also developed, at great expense, a robust system of threat detection and “target hardening” to prevent or defuse actual attacks. An organization called Secure Community Network trains Jewish leaders and community members in situational awareness and self-defense; a rabbi in Texas who was held hostage with three congregants for 11 hours by a jihadist in 2022 credited this training with saving his and his congregants’ lives. Another group, Community Security Initiative, tracks threats on social media 24 hours a day; one flagged online threat to attack synagogues in 2022 led to the arrest in New York’s Penn Station of two men carrying illegal weapons, ammunition, and a swastika armband.

Unfortunately, some bad actors find a sweet spot just past the security cameras. In Los Angeles, harassment of Jews walking to synagogue became common enough in recent years that some formed walking groups with volunteer guards; in December, one street harasser there assaulted an elderly Jewish couple, hitting the husband in the head with a belt buckle, causing a head wound—which was tame compared with a previous incident, in which two Jewish men were shot on their way home from two separate synagogues in February of last year. A week after the belt attack, a man in Washington, D.C., sprayed people leaving a synagogue with what police called a “foul-smelling” substance while shouting “Gas the Jews!”

pro hamas demonstrators

In Minneapolis, a woman who works in communications for a Jewish organization told me how “Free Palestine” had, even before October 7, become a kind of verbal swastika—not because of its meaning, but because of how it is deployed. Apart from its use in political or protest contexts, it has also been used as an online-harassment technique: Trolls tag any post with Jewish content—including material unrelated to Israel—with #FreePalestine, summoning more freedom fighters to the noble cause of verbally abusing Jewish teenagers who dare to post pictures of challah. This verbal vandalism made the jump to real life, the woman explained, and harassers now routinely scrawl it on Jewish communal buildings, shout it at their Jewish schoolmates, and scream it out of car windows at anyone wearing a kippah.

It is remarkable how little any of this has to do with anything going on in the Middle East. This harassment isn’t coming from an antiwar plea, or a consciousness-raising effort about Israeli policies, or a campaign for Palestinian independence, though those pretenses now serve as flimsy excuses. The only purpose of the chalking and swatting and taunting and assaulting and silencing is to dehumanize and demonize Jews. Every time Jews are forced to prove that they didn’t deserve this, or to hide who they are, it is already working.

This new normal for American Jews isn’t just communal, but personal. Many American Jews have quietly dropped friends in recent months after noticing those friends’ posts online casually endorsing the murders of Jews. But even more striking is the low bar for the friends who remain. I’ve seen this most clearly among the young. In upstate New York, a Jewish high schooler told me how a friend of hers regularly passed her cartoons in class. “He just thought it was really funny,” she said, and showed me a sample: a stick-figure caricature of a Hasidic Jew carrying a bag of money. “My friends,” she added, “use my Jewishness to insult me. So they’ll be like, ‘Shut up, you’re just a Jew. Shut up, Jew.’ A couple of my friends say that all the time to me.” I wanted to suggest that she find new friends.

At a Shabbat dinner I attended at one college, students went around the table sharing what they wished they could say to their non-Jewish friends: I wish I could say I want to spend a semester in Israel. I wish I could say I work at a Jewish preschool. I wish I could say I volunteered at a Jewish hospital. I sat at the table stupefied. They were in hiding.

I t was during this ongoing nightmare that Harvard administrators recruited me for advice on the anti-Semitism problem on campus. Against my better judgment, I agreed to join the committee. The Jewish Harvard students who desperately shared their horror stories with me backed them up with piles of evidence. They knew they needed to prove it.

The problem at Harvard, it quickly became clear from the avalanche of documentation deposited at my feet, was not small. The night of the massacre, before the blood was dry, more than 30 Harvard student groups proudly announced that they “hold the Israeli regime entirely responsible for all unfolding violence.” The campus was almost instantly saturated with enthusiastic anti-Israel rallies, which many in the media depicted as the centerpiece of a free-speech debate.

But these protests were not merely outdoor public events that uninterested students could walk past. They also took place inside classroom buildings during lectures, inside the first-year dining hall and inside the largest campus library and other shared study spaces. Jewish students could no longer expect to be able to study in the library, eat in dining halls, or attend class without being repeatedly told by their classmates, sometimes through a bullhorn, that Jews are genocidal murderers deserving of perpetual intifada. (Civilian casualties in war, however horrific, aren’t genocide—but the demonization was the point. So was the vague romanticization of the intifada that targeted, maimed, and murdered Jewish civilians.) At the law school, hundreds of protesters marched through a classroom building during classes. Jewish students reported being targeted and chased through a building by their screaming peers. One video from the business school showed a Jewish student being physically harassed, accosted by protesters who surrounded him with their kaffiyehs.

This demonization of Jews, whether intentional or not, extended to Harvard’s teaching staff. Instructors who grade Jewish students used university-issued class lists to share information about events organized by pro-Palestine groups; at least one even canceled class so students could attend an anti-Israel rally. This pattern among Harvard instructors predated the current Israel-Hamas war. A third-party investigation conducted before the academic year began found that one professor had discriminated against several Israeli students; Harvard said it took action, but the professor rejected the findings and continued teaching. In a separate incident, one student claimed that a different professor asked her to leave his classroom in the spring of 2023 after learning that she was Israeli, because her Israeliness made people “uncomfortable.”

Jewish students who came to Harvard hoping to take courses in Arabic language or Middle Eastern studies told me they often ended up avoiding those courses entirely, wary of professors and peers who made their lack of welcome clear. One recent doctoral student in a field of study unrelated to the Middle East recounted to me that well before October 7, her fellow Ph.D.s in training (the supply pool for teaching assistants) seldom gathered socially without dropping references to “Zionist dirtbags” and “Israeli scum.” One Harvard student described how a classmate, after learning he was Jewish, told him that “there should be no more Jewish state and no more Jews.”

After October 7, social-media platforms exploded with unambiguous Jew hatred in comments such as “Harvard Hillel is burning in hell” and “Let ’em cook.” In this environment, many religious Jewish students stopped wearing kippahs on campus or swapped them for baseball hats; someone spat in the face of one kippah-wearing student as he walked down the street. In an echo of medieval disputations, one Jewish student was invited by a Harvard employee to “debate” him about whether Israel plotted the 9/11 terrorist attacks, according to The Harvard Crimson . Later, the employee posted an online video featuring a screenshot from the student’s X account and the employee wielding a toy machete; the student reported the incident to the authorities and was told to file a restraining order.

Amazingly, Jewish students, whose numbers have dramatically declined at Harvard in recent years for reasons no one seems able to explain, did not respond to all this with their own hate-speech campaigns. Instead, both before and after October 7, Harvard Hillel’s students have reached out to their peers among Harvard’s anti-Israel activists—asking not for a cease-and-desist, but for a dialogue, or even just a cup of coffee. Let’s get to know each other , they offered. The anti-Israel activists refused to engage. Jewish students tried again; they were rebuffed again. And again. This was hardly surprising. For some anti-Israel activists, even merely talking to “Zionists” (a label applied to the 80 percent of American Jews who regard Israel as an essential or important part of their Jewish identity) counts as “normalization”—that is, treating Jews as if they were normal humans, rather than embodiments of evil.

Again we are obliged to prove that this matters. No one died; why complain? “Has there been actual violence against Jewish students at Harvard or on other campuses?” one tenured Harvard professor wrote to our advisory committee to inquire. (The answer was yes.) “If Jewish student worries about physical danger are, in fact, exaggerated,” the professor authoritatively continued, “then students that hold these fears should be advised to leave campus and go home.”

But a hostile environment emerges from pervasive minor incidents, even those that don’t target individuals. Imagine that you are a woman in an office where your male colleagues and bosses gather regularly by the photocopier to discuss their favorite strip clubs. You avoid the photocopier, but then they expand their discussions to the break room, the lobby, the watercooler, the conference room. You avoid those spaces too, avoid those colleagues, hide in your cubicle, and wind up not getting promoted. In such a situation, your company would be responsible for a hostile environment that discriminated against you. The company would not be absolved by pointing out that no one had raped you yet, or that these men weren’t talking to or about you. It could not defend itself by advising you that if these conversations bothered you, you should leave and go home. A hostile environment is precisely one where tenured professors advise students to leave and go home.

The mountain of proof at Harvard revealed a reality in which Jewish students’ access to their own university (classes, teachers, libraries, dining halls, public spaces, shared student experiences) was directly compromised. Compromised, that is, unless they agreed—or at least agreed to pretend, as many Jewish students who are neither religious nor Israeli now silently do—that there was nothing wrong with wallpapering America’s premier university with demonization of Jews. Coercing that silent agreement was the goal, and it was achieved not through arguments or evidence, but through the most laughably idiotic heckler’s veto: screaming at, chasing away, freezing out, or spitting on anyone who dared disagree with supporting the most successful Jew-killers since the Nazis. This left the great minds of Harvard debating the finer points of free speech for hecklers, instead of wondering why their campus was populated by hecklers. The question of why Harvard’s hecklers were heckling in favor of Hamas’s barbarism was too disturbing to consider, and so public discussions ignored it completely.

This heckling was not unrelated to the education that Harvard itself provided. Classes existed at Harvard, it turned out, that were premised on anti-Semitic lies. A course at the school of public health called “The Settler Colonial Determinants of Health” looked at case studies from South Africa, the United States, and Israel; its premise—not a topic of discussion, but the premise on which the course was built—was that Israel is a settler-colonialist state. (A Jewish student who wrote to the professor questioning what they saw as the ideological slant of the readings was told that it was “insulting” to suggest that the course had an agenda.) The “Palestine Program for Health and Human Rights” proudly announced that it “utilizes a decolonial framework in program development, leadership, and engagement”—meaning, one might reasonably assume, the “decolonizing” of Israel through the removal of its 7 million Jews. (The program is a partnership between Harvard and Birzeit University, a Palestinian institution where an Israeli journalist was expelled from an event in 2014 just because she was Israeli and Jewish.)

An astonishing number of pop-up lectures, panels, and events at Harvard both before and after October 7 were centered on the suffering of Palestinians in Gaza—a worthy topic addressed with almost no mention of Hamas, even though Hamas has ruled Gaza for 17 years. Nor was there much mention of the fact that Hamas was founded in connection with the global Muslim Brotherhood, or of its comically wealthy sponsors in the Persian Gulf. Students had many opportunities to learn about Palestinian suffering from oppression by evil Jews, but far fewer opportunities to learn, for instance, about Hamas’s success in co-opting foreign aid and crushing dissent, or the intifada that students hoped to globalize. Outside of their engagements at Harvard, some guest speakers publicly endorsed extreme anti-Semitic lies, including the straight-up blood libel that Israelis are harvesting Palestinians’ organs or that the Israeli military uses Palestinian children for weapons testing. One could hardly blame students for repeating their educators’ claims.

Hillary Rodham Clinton: Hamas must go

Out of respect for Gay’s request that our committee’s discussions with administrators remain private, I won’t share here anything that we talked about in our many meetings. But I will say that one thing we did not discuss was Gay’s congressional testimony on this topic, for which she and other administrators never asked for the advisory committee’s advice. Instead, they consulted lawyers, a choice that backfired on national television.

The horror that the hearing laid bare was something far worse than a viral gaffe. Harvard was already being investigated by the Department of Education for allegations of violating Jewish students’ civil rights under Title VI, and perhaps the president was advised against admitting any institutional failure. (In January, a group of students sued Harvard, describing the university as a “bastion of rampant anti-Jewish hatred and harassment.”) Still, the only morally tenable position would have been to admit failure, to reveal that the problem was not all in Jews’ heads; that there truly was an anti-Semitic environment at these incubators of American leadership; that these universities, along with far too many other pockets of the country, had reverted, slowly and then all at once, into what they had been a century earlier: safe spaces for high-minded Jew hatred—not in spite of their aspiration that education should lead to a better world, but because of it.

I t is fairly obvious what Harvard and other universities would need to do to turn this tide. None of it involves banning slogans or curtailing free speech. Instead it involves things like enforcing existing codes of conduct regarding harassment; protecting classroom buildings, libraries, and dining halls as zones free from advocacy campaigns (similar to rules for polling places); tracking and rejecting funding from entities supporting federally designated terror groups (a topic raised in recent congressional testimony regarding numerous American universities); gut-renovating diversity bureaucracies to address their obvious failure to tackle anti-Semitism; investigating and exposing the academic limitations of courses and programs premised on anti-Semitic lies; and expanding opportunities for students to understand Israeli and Jewish history and to engage with ideas and with one another. There are many ways to advocate for Israeli and Palestinian coexistence that honor the dignity and legitimacy of both indigenous groups and the need to build a shared future. The restoration of such a model of civil discourse, which has been decimated by heckling and harassment, would be a boon to all of higher education.

Harvard has already begun signaling change in this direction: The university recently reiterated and clarified rules regarding the time, place, and manner of student protests. For Harvard to take more of these steps would be huge, but I have struggled to understand why all of them still feel so small. Perhaps it’s because the problem is a multi-thousand-year fatal flaw in the ways our societies conceive of good and evil—and also because somewhere deep within me, I know what has been lost. There was a time, not so very long ago, when we didn’t have to prove our right to exist.

Among the mountains of evidence that Jewish students sent me, one image has stayed in my mind. There are videos of crowds chanting “Long live the intifada!” inside Harvard’s Science Center, and “There is only one solution: intifada revolution!” in Harvard Yard, along with other places equally familiar from my student days. But I keep coming back to the crowds marching and screaming in front of Harvard Law School’s Langdell library, because Langdell is a sacred place for me. On my 22nd birthday, in 1999, when I was a senior at Harvard, a law student I’d met at Hillel took me up through Langdell’s maintenance passageways to the library’s rooftop, where he asked me to marry him. I said yes.

I watched the video of the students marching and screaming in front of Langdell, and in an instant I remembered everything: studying in campus libraries for my Hebrew- and Yiddish-literature courses, talking for hours with Muslim and Christian and progressive and conservative classmates, inviting friends of all backgrounds to join me at Hillel, scrupulously following the Jewish tradition of “argument for the sake of heaven” in even the most heated debates, gathering for Shabbat dinners crowded with hundreds of students—and over those long and beautiful dinners, falling in love. My classmates and I often disagreed about the most important things. But no one screamed in our faces when we wore Hebrew T-shirts on campus. No one shunned us when we talked about our friends and family in Israel, or spat on us on our way to class. No crowds gathered to chant for our deaths. No one told us that there should be no more Jews. That night, my future husband and I worried only about getting in trouble for sneaking up to the library roof.

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Bring Human Values to AI

  • Jacob Abernethy,
  • François Candelon,
  • Theodoros Evgeniou,
  • Abhishek Gupta,
  • Yves Lostanlen

free fall problem with solution

When it launched GPT-4, in March 2023, OpenAI touted its superiority to its already impressive predecessor, saying the new version was better in terms of accuracy, reasoning ability, and test scores—all of which are AI-performance metrics that have been used for some time. However, most striking was OpenAI’s characterization of GPT-4 as “more aligned”—perhaps the first time that an AI product or service has been marketed in terms of its alignment with human values.

In this article a team of five experts offer a framework for thinking through the development challenges of creating AI-enabled products and services that are safe to use and robustly aligned with generally accepted and company-specific values. The challenges fall into five categories, corresponding to the key stages in a typical innovation process from design to development, deployment, and usage monitoring. For each set of challenges, the authors present an overview of the frameworks, practices, and tools that executives can leverage to face those challenges.

Speed and efficiency used to be the priority. Now issues such as safety and privacy matter too.

Idea in Brief

The problem.

Products and services increasingly leverage artificial intelligence to improve efficiency and performance, but the results can be unpredictable, intrusive, offensive, and even dangerous.

The Solution

Companies need to factor AI’s behavior and values into their innovation and development processes to ensure that they bring to market AI-enabled offerings that are safe to use and are aligned with generally accepted and company-specific values.

How to Proceed

This article identifies six key challenges that executives and entrepreneurs will face and describes how to meet them. Companies that move early to acquire the needed capabilities will find them an important source of competitive advantage.

When it launched GPT-4, in March 2023, OpenAI touted its superiority to its already impressive predecessor, saying the new version was better in terms of accuracy, reasoning ability, and test scores—all of which are AI-performance metrics that have been used for some time. However, most striking was OpenAI’s characterization of GPT-4 as “more aligned”—perhaps the first time that an AI product or service has been marketed in terms of its alignment with human values.

  • JA Jacob Abernethy is an associate professor at the Georgia Institute of Technology and a cofounder of the water analytics company BlueConduit.
  • FC François Candelon is a managing director and senior partner at Boston Consulting Group (BCG), and the global director of the BCG Henderson Institute.
  • Theodoros Evgeniou is a professor at INSEAD and a cofounder of the trust and safety company Tremau.
  • AG Abhishek Gupta is the director for responsible AI at Boston Consulting Group, a fellow at the BCG Henderson Institute, and the founder and principal researcher of the Montreal AI Ethics Institute.
  • YL Yves Lostanlen has held executive roles at and advised the CEOs of numerous companies, including AI Redefined and Element AI.

free fall problem with solution

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AT&T’s network is having problems: What you should know while navigating a phone service outage

AT&T says it has restored wireless coverage after an outage knocked out cellphone service on its network across the U.S. for hours. (Feb. 22) (AP video by Kendria LaFleur)

FILE _ A cellular phone tower is shown on Monday, May 22, 2017 in High Ridge, Mo. A number of Americans are dealing with cellular outages Thursday, Feb. 22, 2024, on AT&T, Cricket Wireless, Verizon, T-Mobile and other service providers, according to data from Downdetector. (AP Photo/Jeff Roberson)

FILE _ A cellular phone tower is shown on Monday, May 22, 2017 in High Ridge, Mo. A number of Americans are dealing with cellular outages Thursday, Feb. 22, 2024, on AT&T, Cricket Wireless, Verizon, T-Mobile and other service providers, according to data from Downdetector. (AP Photo/Jeff Roberson)

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FILE - A woman looks at her phone while watching the sun set as triple-digit heat indexes continue in the Midwest Sunday, Aug. 20, 2023, in Kansas City, Mo. A number of Americans on Thursday, Feb. 22, 204, are dealing with cellular outages on AT&T, Cricket Wireless, Verizon, T-Mobile and other service providers, according to data from Downdetector. AT&T, who was the hardest hit, is actively working to restore service to all of its customers. (AP Photo/Charlie Riedel, File)

FILE -This July 18, 2019, file photo, shows an AT&T retail store in Miami. Customers of AT&T, the country’s largest wireless provider, reported widespread outages on Thursday, Feb. 22, 2024. (AP Photo/Lynne Sladky, File)

FILE - In this Friday, April 30, 2021 file photo, a man uses a mobile phone in New York. A number of Americans on Thursday, Feb. 22, 204, are dealing with cellular outages on AT&T, Cricket Wireless, Verizon, T-Mobile and other service providers, according to data from Downdetector. AT&T, who was the hardest hit, is actively working to restore service to all of its customers. (AP Photo/Mark Lennihan, File)

NEW YORK (AP) — Customers of AT&T, the country’s largest wireless provider, reported widespread outages on Thursday .

Here’s what to know if you are having problems with your phone service.

When did the outage start?

AT&T had more than 58,000 outages around noon ET , in locations including Houston, Atlanta and Chicago. The outages, which began at approximately 3:30 a.m. ET, peaked at around 73,000 reported incidents. The carrier has more than 240 million subscribers, the country’s largest.

FILE - Nokia logo seen in the Mobile World Congress 2023 in Barcelona, Spain, Tuesday, Feb. 28, 2023. Nokia on Thursday reported a double-digit decline in sales and a fall in profit in the last three months of 2023, with the wireless and fixed-network equipment maker saying operators are cutting back on investments into 5G and other technology because of economic uncertainty. (AP Photo/Joan Mateu Parra, File)

What is the SOS mode?

Some AT&T iPhone customers saw SOS messages displayed in the status bar on their cellphones. The message indicates that the device is having trouble connecting to their cellular provider’s network, but it can make emergency calls through other carrier networks, according to Apple Support.

What is Wi-Fi calling?

AT&T urged customers to connect to Wi-Fi to use their phones. Wi-Fi calling is a built-in feature on most Android devices and iPhones and can be turned on under the phone’s settings.

“Some of our customers are experiencing wireless service interruptions this morning. We are working urgently to restore service to them. We encourage the use of Wi-Fi calling until service is restored,” AT&T said in a statement.

If Wi-Fi isn’t available, there are few options for cell phone users. It’s possible to switch services if a phone is unlocked, but that requires signing up online and porting your phone number .

Some apps, including Google Maps, have limited service offline. Payment apps also do not use a phone’s cell service to work and should also be useable.

What caused the outage?

AT&T has stayed mum so far on what caused the outage.

free fall problem with solution

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A new mother holds her baby wearing a yellow jumper

Birthrate in UK falls to record low as campaigners say ‘procreation a luxury’

Total fertility was 1.49 children per woman in 2022 amid rising housing and childcare costs

Campaigners have warned that “procreation has become a luxury item”, after it emerged that the fertility rate in England and Wales had fallen to its lowest level since records began in 1939.

Official figures from the Office for National Statistics (ONS) showed “total fertility”, calculated based on the birthrate across different age groups, fell to 1.49 children per woman in 2022.

That is well below the rate of 2.1 needed to maintain a steady population without significant immigration. In total, there were 605,479 live births in 2022, according to the ONS, down 3.1% from a year earlier, and the lowest number since 2002.

Falling birthrates since 2010 have already prompted schools closures in many areas in recent years, including central London.

Dr Mary-Ann Stephenson of the Women’s Budget Group, which campaigns for more support for families, said: “We need the babies who are born now, because they will be the people whose taxes pay for our healthcare. These will be the people looking after us in our old age. These will be the doctors and nurses and care workers of the future.”

Stephenson said tackling the issue was not about persuading people to become parents, but “making sure that we have systems in place to support those people who want children to have children”. She pointed to unaffordable housing as a significant factor.

Campaigners also warned that rocketing childcare costs were likely to have contributed to some women’s decision not to have children – or to have fewer children than they would have liked.

Joeli Brearley, chief executive of Pregnant Then Screwed, said: “It is no surprise to us that fertility rates have hit the floor. Procreation has become a luxury item in the UK. Childcare costs are excruciating, and that’s if you can secure a place.

“Our research found that almost half of parents have been plunged into debt or had to use savings just to pay their childcare bill,” she added.

Phoebe Arslanagić-Little, head of the new deal for parents at the conservative thinktank Onward, said: “We are essentially pricing people out of parenthood, with a panoply of material issues: the housing crisis, childcare costs, the poor availability of IVF on the NHS.”

She called for more support for parents in the tax and welfare systems, adding: “We don’t do a good enough job of reflecting the actual contribution that parents are making”.

The wider economic climate is also known to be a factor. Research published by the Bank of England in 2020 suggested that the sharp reduction in interest rates at the time of the global financial crisis may have led to as many as 15,000 extra babies being born the following year because households were handed a financial windfall.

The ONS data also showed that women are tending to have children later: the fertility rate was highest among women aged 30-34, whereas before 2002 it was higher in the 25-29 age group.

Fertility rates have also been falling across much of Europe in recent years, with women having fewer children, and having them later in life.

Politicians in the UK have tended to be wary of encouraging people to have more children, for fear of being seen to interfere in voters’ private lives.

But some Conservatives have recently raised the alarm. Miriam Cates, a backbench MP, warned last year that falling birthrates were “the one overarching threat to British conservatism and to the whole of western society”.

  • Office for National Statistics
  • UK cost of living crisis

More on this story

free fall problem with solution

Humans could face reproductive crisis as sperm count declines, study finds

free fall problem with solution

Total fertility rate rises for first time in a decade in England and Wales

free fall problem with solution

Air pollution may affect sperm quality, says study

free fall problem with solution

Record numbers of women reach 30 child-free in England and Wales

free fall problem with solution

‘I’m scared I’ve left it too late to have kids’: the men haunted by their biological clocks

free fall problem with solution

Women are still being blamed for society’s problems with fertility

free fall problem with solution

Fall in fertility rates may be linked to fossil fuel pollution, finds study

free fall problem with solution

Australia’s fertility rate falls to record low in 2020

free fall problem with solution

Don’t blame women for our low birthrate – we need to fix our precarious society

free fall problem with solution

Italy’s birthrate is falling. Can the storks help?

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IMAGES

  1. Free Fall Problems

    free fall problem with solution

  2. Free Fall :: Physics Tutorials

    free fall problem with solution

  3. HTPIB02H Free Fall Problems Example 2

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  4. Free Fall Motion

    free fall problem with solution

  5. free-fall-practice-problems-1

    free fall problem with solution

  6. Physics Free Fall Practice Problem 3

    free fall problem with solution

VIDEO

  1. HTPG02F Free Fall Problem Solving

  2. Does free fall depends on mass ? #shorts #scienceexperiment #physics

  3. Pet hair fall problem solution

  4. Free Fall

  5. Cutnell and Johnson 9e Chapter 2 Problem 52

  6. Solving a Harder Free Fall Problem

COMMENTS

  1. Free Fall Problems

    Free Fall Problems On this page I put together a collection of free fall problems to help you understand the concept of free fall better. The required equations and background reading to solve these problems are given here, for θ = 90°. Problem # 1 A ball is thrown with an initial upward velocity of 5 m/s.

  2. Free fall formula physics

    Answer: The Velocity in free fall is autonomous of mass. V (Velocity of iron) = gt = 9.8 m/s 2 × 5s = 49 m/s V (Velocity of cotton) = gt = 9.8 m/s 2 × 3s = 29.4 m/s. The Velocity of iron is more than cotton. Freefall is a body falling freely because of the gravitational pull of our earth. Freefall formulas and related examples.

  3. Free Fall (Physics): Definition, Formula, Problems & Solutions (w

    Updated December 22, 2020 By Kevin Beck Free fall refers to situations in physics where the only force acting on an object is gravity. The simplest examples occur when objects fall from a given height above the surface of the Earth straight downward - a one-dimensional problem.

  4. 3.5 Free Fall

    Gravity The most remarkable and unexpected fact about falling objects is that if air resistance and friction are negligible, then in a given location all objects fall toward the center of Earth with the same constant acceleration, independent of their mass.

  5. Kinematic Equations and Free Fall

    Applying Free Fall Concepts to Problem-Solving There are a few conceptual characteristics of free fall motion that will be of value when using the equations to analyze free fall motion. These concepts are described as follows: An object in free fall experiences an acceleration of -9.8 m/s/s. (The - sign indicates a downward acceleration.)

  6. Solving freefall problems using kinematic formulas

    Google Classroom You might need: Calculator A squirrel drops an acorn onto the head of an unsuspecting dog. The acorn falls 4.0 m before it lands on the dog. We can ignore air resistance. How many seconds did the acorn fall? Round the answer to two significant digits. Answer using a coordinate system where upward is positive. s Show Calculator

  7. Free Fall Motion: Tutorials with Examples and Solutions

    Problems on free fall motion are presented along with detailed solutions. Problem 1: From rest, a car accelerated at 8 m/s 2 for 10 seconds. a) What is the position of the car at the end of the 10 seconds? b) What is the velocity of the car at the end of the 10 seconds? Solution to Problem 1 Problem 2:

  8. Solving Free Fall Problems (with 5 Examples)

    Difficulty solving free fall problems doesn't have to be your downfall. We can help. This video springboards off of two other videos - our Describing Free Fa...

  9. 3.7: Free Fall

    An interesting application of Equation 3.3.2 through Equation 3.5.22 is called free fall, which describes the motion of an object falling in a gravitational field, such as near the surface of Earth or other celestial objects of planetary size.Let's assume the body is falling in a straight line perpendicular to the surface, so its motion is one-dimensional.

  10. Free Fall

    The first is "drop mode" — a straight free fall. During a simple drop experiment, the capsule is pulled up to a height of 120 meters to the top of the drop tube and then released. After 4.74 seconds the experiment has landed safely in the deceleration unit filled with polystyrene pellets.

  11. Free Fall Physics Problems

    This physics video tutorial focuses on free fall problems and contains the solutions to each of them. It explains the concept of acceleration due to gravity...

  12. 4.6: Free Fall

    4.6: Free Fall. Use the kinematic equations with the variables y and g to analyze free-fall motion. Describe how the values of the position, velocity, and acceleration change during a free fall. Solve for the position, velocity, and acceleration as functions of time when an object is in a free fall.

  13. Free Fall Motion: Explanation, Review, and Examples

    Last Updated On: February 16, 2023 Free fall and projectile motion describe objects that are moving through the air and acted on only by gravity. In this post, we will describe this type of motion using both graphs and kinematic equations. Since projectile motion involves two dimensions, these problems can be complex.

  14. Free Fall :: Physics Tutorials

    We answer these questions now. Picture given above shows the magnitudes of velocity at the bottom and at the top. As you can see the ball is thrown upward with an initial v velocity, at the top it's velocity becomes zero and it changes it's direction and starts to fall down which is free fall.

  15. Free fall motion

    Free fall motion - problems and solutions 1. A stone free fall from the height of 45 meters. If the acceleration due to gravity is 10 ms-2, what is the speed of the stone when hits the ground? Known : Height (h) = 45 meters Acceleration due to gravity (g) = 10 m/s2 Wanted : The final velocity of the stone when it hits the ground (vt) Solution :

  16. Free Fall

    practice problem 1. The following passages are excerpts from "The Long, Lonely Leap" by Captain Joseph Kittinger USAF as they appeared in National Geographic magazine. It is the story of his record-setting, high altitude parachute jump from a helium balloon over New Mexico on 16 August 1960.

  17. Free Fall

    D vs t - free fall . A ball is thrown upward. What's the acceleration due to gravity at the peak? -9.8 m/s 2 . AP Question . Ex) Compare the displacement of an object dropped during its first second with its displacement after 3 seconds. Δd = V i t + 1/2aΔt 2 . Δd = 1/2aΔt 2 . 9 times greater

  18. Free Fall Practice Problems for High Schools: Complete Guide

    Solution: Take up as the positive direction and the throwing point as the origin, so y_0=0 y0 = 0 . (a) The ball goes up so high that its vertical velocity becomes zero. For this part of ascending motion, we can use the free fall kinematic equation v^2-v_0^2=-2g (y-y_0) v2 −v02 = −2g(y − y0).

  19. PDF Example 8.12 Free Fall with Air Drag

    Example 8.12 Free Fall with Air Drag. m that is in free fall but experiencing air resistance. The is given by Eq. (8.6.1), where ρ is the density of air, A is the cross-sectional area of the object in a plane perpendicular to the motion, and C is. the drag coefficient. Assume that the object is released from rest and very quickly attains ...

  20. 1.1: Free Fall

    Home Bookshelves Differential Equations A First Course in Differential Equations for Scientists and Engineers (Herman) 1: First Order ODEs

  21. Free fall

    Solving freefall problems using kinematic formulas Worked example: Free fall, object thrown up from a building Advanced: Freefall problems Science > Mechanics (Essentials) - Class 11th > How to analyze car crashes using analysis of skid marks? > Why does a feather and a steel ball fall at the same time in freefall! © 2024 Khan Academy Terms of use

  22. PDF Section 3 Free Fall: Practice Problems

    41. A construction worker accidentally drops a brick from a high scaffold. What is the velocity of the brick after 4.0 s? How far does the brick fall during this time? SOLUTION: Let upward be the positive direction. The brick falls 78 m. ANSWER: Let upward be the positive direction. = 39 m/s downward f x = -78 m

  23. Free Fall Calculator

    Calculate the final free fall speed (just before hitting the ground) with the formula: v = v₀ + gt = 0 + 9.80665 × 8 = 78.45 m/s. Find the free fall distance using the equation: s = (1/2)gt² = 0.5 × 9.80665 × 8² = 313.8 m. If you know the height from which the object is falling, but don't know the time of fall, you can use this ...

  24. The U.S.-Mexico Border: How Americans View the Situation, Its Causes

    Democrats mostly view the situation as a major problem (44%) or minor problem (26%) for the U.S. Very few Democrats (7%) say it is not a problem. In an open-ended question, respondents voice their concerns about the migrant influx. They point to numerous issues, including worries about how the migrants are cared for and general problems with ...

  25. The Return of the Big Lie: Anti-Semitism Is Winning

    The problem was not that Jewish students on American university campuses didn't want free speech, or that they didn't want to hear criticism of Israel. Instead, they didn't want people ...

  26. Bring Human Values to AI

    The Problem. Products and services increasingly leverage artificial intelligence to improve efficiency and performance, but the results can be unpredictable, intrusive, offensive, and even dangerous.

  27. AT&T outage: What to do when your phone service is down

    Here's what to know if you are having problems with your phone service. When did the outage start? AT&T had more than 58,000 outages around noon ET, in locations including Houston, Atlanta and Chicago. The outages, which began at approximately 3:30 a.m. ET, peaked at around 73,000 reported incidents. The carrier has more than 240 million ...

  28. Birthrate in UK falls to record low as campaigners say 'procreation a

    Record numbers of women reach 30 child-free in England and Wales ... Women are still being blamed for society's problems with fertility. 13 Oct 2021. Fall in fertility rates may be linked to ...