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Lecture Comments (68)

1 answer

Last reply by: Professor Dan Fullerton
Sat Apr 22, 2017 4:29 PM

Post by Moksh Modi on April 22 at 04:16:45 PM

Hi Professor Dan. Really like all your lectures. I would just like to know how can you memorize the projectile Motion formulas for the test.

1 answer

Last reply by: Professor Dan Fullerton
Tue Mar 28, 2017 6:55 AM

Post by Parsa Abadi on March 27 at 07:21:53 PM

What does above the horizontal line mean? above ground level?

1 answer

Last reply by: Professor Dan Fullerton
Wed Jan 18, 2017 5:54 AM

Post by Elman Ahmed on July 7, 2016

It's amazing how you organized and kept all the information so nicely. I Never saw anyone doing projectile motions in such an organized way and same thing goes to collision problem. I really liked how you approached Collision and projectile motion problems. Very ORGANIZED indeed!

1 answer

Last reply by: Dukaiwen Zhao
Tue Jul 5, 2016 3:14 PM

Post by Dukaiwen Zhao on July 5, 2016

For example 5, why delta y = -8?

2 answers

Last reply by: Muhammad Razzaq
Tue Jan 17, 2017 3:34 PM

Post by Jim Tang on October 18, 2015

in some of the examples, there was a horizontal force acting on the projectile throughout its flight. so technically the only force acting on it was not the force of gravity..?

1 answer

Last reply by: Professor Dan Fullerton
Wed Jan 18, 2017 5:56 AM

Post by Jim Tang on October 18, 2015

around 15:25, on the bottom velocity and acceleration for y,

if you take the area under the acceleration up to the time the velocity hits 0, i don't think the area under the acceleration curve is 0. doesn't this break the rule?

2 answers

Last reply by: Bilbo Baggins
Tue Sep 1, 2015 6:59 PM

Post by Bilbo Baggins on September 1, 2015

I don't believe the equation in example 1 was covered in the previous lesson, that equation is :  delta x = v * t


2 answers

Last reply by: Bilbo Baggins
Tue Sep 1, 2015 6:47 PM

Post by Bilbo Baggins on August 31, 2015

why are the kinematic equations for projectile motion not included here?

1 answer

Last reply by: Professor Dan Fullerton
Fri Jul 24, 2015 6:55 AM

Post by Jim Tang on July 24, 2015

In Example #5, we have three items in for the horizontal component, so why didn't we use the rule that if we have 3, we can solve it all? I feel the time is very important to projectile motion for both horizontal and vertical components?

In addition, how, in general, are we applying scientific motion to every problem? I've seen problems were we start with 1 sig fig and end up with more than 1 sig fig.

Terrific videos!

2 answers

Last reply by: Anh Dang
Tue Jul 21, 2015 1:18 PM

Post by Anh Dang on July 20, 2015

In the second to last example (Launched from a height), where do you get -10 for the Ay?

1 answer

Last reply by: Professor Dan Fullerton
Fri Jan 30, 2015 4:38 AM

Post by Muhammad Chauhan on January 29, 2015

Thank you for your lectures.

In example 4, X position vs time graph shows that the x value is increasing infinitely... How can this be correct when we eventually stop moving in the x direction?

1 answer

Last reply by: Professor Dan Fullerton
Sun Dec 28, 2014 12:36 AM

Post by Micheal Bingham on December 27, 2014

Hi, since the only force behaving on an object thrown upward is gravity, why is it that it pauses in mid air for a brief second before falling back down?

1 answer

Last reply by: Daniel Fullerton
Tue Nov 4, 2014 6:38 AM

Post by David Millaud on November 4, 2014

This seems to be extremely confusing the way u have set this problem up because the Vy in your case works out to be 149.5.  If we understand that max height can be determine by avg. velocity multiplied by time.  An that time can be determine by dividing initial velocity in the vertical direction by acceleration due to gravity.  Then we will get the max height for v=0.  Once determine we can use the equation ^y=Voy(t)+1/2at^2= 1125-8+0+1/2(-10)(t)^2 will work out for the time at max height to check our work.  Mulitply that time by two for total time and then multiply it by the Vox we will have max distance in horizontal direction.  Also i don't know if that equation you used will suffice because the object although is moving in a parabolic fashion it is not returning to the same ground level.  According to the rules for that equation.

1 answer

Last reply by: Professor Dan Fullerton
Mon Oct 20, 2014 3:37 PM

Post by Sally Acebo on October 20, 2014

In ex5, why is delta y =-8m? How did it become negative? It is going up so wouldn't be positive?

1 answer

Last reply by: Professor Dan Fullerton
Thu Oct 2, 2014 5:55 AM

Post by Derek Boutin on October 1, 2014

In all of these examples, would mass and air resistance affect the answers if we chose not to neglect them?

1 answer

Last reply by: Professor Dan Fullerton
Wed Oct 1, 2014 6:21 AM

Post by Derek Boutin on September 30, 2014

In Example 1, wouldn't the final horizontal velocity be 0 if we assume it hits the ground?

1 answer

Last reply by: Professor Dan Fullerton
Thu Sep 4, 2014 2:04 PM

Post by Okwudili Ezeh on September 4, 2014

I came across this question and the answer given was 2%, but I do not know how they came about this answer.

A projectile is fired vertically at a speed of 30.0m/s. It reaches a maximum height of 44.1 m. What fraction of its initial energy has been lost to air resistance at this point?

1 answer

Last reply by: Professor Dan Fullerton
Fri Jun 27, 2014 8:46 PM

Post by robert moreno on June 23, 2014

You're awesome!

3 answers

Last reply by: Professor Dan Fullerton
Tue Nov 4, 2014 6:39 AM

Post by Andrew Ablett on April 16, 2014

Hi there Dan, great lecture.

For example 5 I get the total distance of the parabola to be 7,800m... but since the projectile is fired 8m above the water, don't we need to add the distance traveled within that extra 8m of free fall. In which case I get 7, 800 + 327.6 (the distance of the canon ball in the x direction in 1.26 seconds of free fall)= 8, 127.6m for when the canon ball hits the water.

Please let me know where I have gone wrong. Thank you.  

0 answers

Post by Professor Dan Fullerton on February 17, 2014

We're trying to find DeltaX, so that's unknown.  We know the y displacement has to be 2m down since it starts 2 m above the ground, and its path ends when it is on the ground, a change in y of 2 meters.

0 answers

Post by rumeh mandible on February 17, 2014

in this example why does displacement y has 2 and displace x has non

1 answer

Last reply by: Professor Dan Fullerton
Sun Jan 12, 2014 3:36 PM

Post by Seetha Mahadevan on January 12, 2014

In the first example how did you find that the vertical acceleration equaled 10 m/s^2?

1 answer

Last reply by: Professor Dan Fullerton
Fri Dec 20, 2013 7:36 AM

Post by Hyun Cho on December 19, 2013

in the first example, why isnt the final velocity 0 when it hits the ground? since it wont move once it reaches the ground, isnt the velocity 0?

1 answer

Last reply by: Professor Dan Fullerton
Mon Oct 7, 2013 6:17 AM

Post by chisom linda madumere on October 6, 2013

in example 5 y is the change in y= -8.y the negative sign

1 answer

Last reply by: Professor Dan Fullerton
Mon Oct 7, 2013 6:18 AM

Post by Yuen Ting Wong on October 6, 2013

I finally get what it is ! You are much better than the one teaching me in school!! Thank you~~

0 answers

Post by Professor Dan Fullerton on September 15, 2013

Because you have to break the initial velocity into its horizontal and vertical components.  If the total initial velocity is 300 m/s, and it's launched at an angle of 30 degrees above the horizontal, the vertical component of the initial velocity is 300m/s*sin(30)=150 m/s.

1 answer

Last reply by: Professor Dan Fullerton
Mon Oct 7, 2013 6:18 AM

Post by Constance Kang on September 15, 2013

in the example 5,why is the initial velocity of y 150m/s instead of 300m/s?

1 answer

Last reply by: Professor Dan Fullerton
Tue Sep 10, 2013 6:25 PM

Post by Jennifer Salcedo on September 10, 2013

In example 5, I solved for time by using delta x is equal to final velocity times time & for .053 seconds, then I solved for time again using delta x is equal to initial velocity times time plus .5a(t^2) and got .053 seconds again, I'm really confused.

3 answers

Last reply by: Professor Dan Fullerton
Tue Jul 2, 2013 10:54 AM

Post by Mimi Nguyen on July 2, 2013

Thank you so much for your help in physics, Professor Fullerton.

For example 3 with the Human Cannonball, the time was multiplied by two because the vertical was the midpoint and the parabola is symmetrical. However, in example 5, the time wasn't doubled even though a cannonball was launched instead of dropped as well. Could you please clarify when and when not to double the time?

Thank you in advance.

1 answer

Last reply by: Professor Dan Fullerton
Mon Jun 24, 2013 9:38 AM

Post by Javier Torres on June 23, 2013

In example 5, why was delta y set at -8m?

2 answers

Last reply by: Professor Dan Fullerton
Mon Mar 25, 2013 6:53 PM

Post by Ikze Cho on March 25, 2013

in a scenario where a ball is thrown:
Is the final velocity always equal to 0?

Related Articles:


  • Projectiles are objects acted upon only by gravity.
  • Projectiles launched at angles move in parabolic arcs.
  • Vertical motion and horizontal motion are completely independent. Horizontal motion does not affect vertical motion, and vice versa.
  • An object will travel the maximum horizontal distance across level ground with a launch angle of 45 degrees.


Lecture Slides are screen-captured images of important points in the lecture. Students can download and print out these lecture slide images to do practice problems as well as take notes while watching the lecture.

  • Intro 0:00
  • Objectives 0:06
  • What is a Projectile? 0:26
    • An Object That is Acted Upon Only By Gravity
    • Typically Launched at an Angle
  • Path of a Projectile 1:03
    • Projectiles Launched at an Angle Move in Parabolic Arcs
    • Symmetric and Parabolic
    • Horizontal Range and Max Height
  • Independence of Motion 2:17
    • Vertical
    • Horizontal
  • Example 1: Horizontal Launch 3:49
  • Example 2: Parabolic Path 7:41
  • Angled Projectiles 8:30
    • Must First Break Up the Object's Initial Velocity Into x- and y- Components of Initial Velocity
    • An Object Will Travel the Maximum Horizontal Distance with a Launch Angle of 45 Degrees
  • Example 3: Human Cannonball 8:55
  • Example 4: Motion Graphs 12:55
  • Example 5: Launch From a Height 15:33
  • Example 6: Acceleration of a Projectile 19:56

Transcription: Projectiles

Hi everybody and welcome back to

In this lesson we are going to talk about projectiles.0003

Our objectives are going to be to sketch the theoretical path of a projectile, to recognize the independence of the vertical and horizontal motions of a projectile, and to solve problems involving projectile motion for projectiles fired horizontally and projectiles fired at an angle.0006

Let us dive right in.0026

What is a projectile?0027

A projectile is an object that is acted upon only by gravity.0029

In reality, air resistance plays a role -- we all know it does, but for the purposes of this analysis we are going to neglect air resistance.0034

Typically, projectiles are objects launched at an angle.0043

So a projectile is probably something that gets launched up in the air, comes back down or is launched, comes back down horizontally, or is launched down and travels that direction -- typical projectile.0046

Let us start by talking about the path of the projectile.0062

Projectiles launched at an angle move in parabolic arcs.0065

Let us draw our ground here.0070

If we launch a projectile with some initial velocity, let us call that V-initial, at some angle theta, it is going to travel up in a parabolic arc and come back down.0073

And that should be symmetric other than I have a pretty poor writing.0088

It is symmetric and it is parabolic.0091

It is coming back down to the same height it was launched from -- it will travel up in an arc and back down -- split it in two and you should have a mirror image of each other.0099

Now the distance it travels along the ground is known as its horizontal range.0108

We could also look at the max height that the projectile reaches.0116

And one key, if you want a projectile launched from level ground to go the furthest possible distance, the angle you want to launch it at -- 45 degrees. Okay?0123

Let us take a look at the independence of motion.0136

Projectiles, objects launched in two dimensions -- it is the same physics we have been doing. They are easy.0140

The same thing we have been doing with kinematic equations, we are just going to do it again.0148

The only difference is we are going to treat the vertical aspect of the projectile and the horizontal aspect separately.0151

As long as we keep them separate, we can keep doing all the same math we have been doing.0160

No real new analysis needs to be brought into the fold.0165

Vertical is just like the free fall problems -- and horizontal -- even simpler. There is no acceleration.0169

Gravity never pulls you sideways -- does not happen. Gravity only pulls down.0174

So as long as we treat them separately our analyses get pretty easy.0180

What we are going to have to remember though, is breaking vectors up into components.0184

If for example, we have an initial velocity for a vector -- some initial velocity launched at some angle theta -- we need to simplify it.0188

We need to make this easy by breaking it up into an X component -- V initial x -- which hopefully you remember is V initial cos theta and V initial y component -- V initial y, which is V initial sin of the angle theta.0199

Then we are going to deal with this V initial x and this V initial y vector separately.0220

Let us see how that works.0227

Fred throws a baseball 42 m/s horizontally from a height of 2 m.0230

How far will the ball travel before it reaches the ground?0234

Well, a great place to start -- drawing a diagram.0238

There is our ground -- we have Fred, who is launching the baseball 2 m from the ground.0242

There is our baseball, and we know it is going to follow something of a parabolic arc like that and its initial velocity, V theta is, pardon me, 42 m/s.0251

All right, so to do our analysis, we are going to look at the horizontal and vertical aspects separately again.0268

Horizontally, let us call to the right the positive x direction.0273

We have V initial x, V final x, delta x, Ax and T.0279

The initial velocity in the x direction is 42 m/s.0289

The final velocity in the x direction has to be the same because gravity does not accelerate things sideways, only down.0293

Acceleration in the x direction is 0.0302

All right, let us make our table for what we know vertically.0307

Vertically, our object is going to go down first so let us call down the positive y direction.0311

We have V initial y, V final y, delta y, Ay and T.0317

V initial y for an object launched horizontally, if all of its velocity is this way, there is no initial vertical component. That is 0.0326

We do not know what its final vertical component is going to be but we know its displacement vertically is going to be 2 m because that is how far it travels vertically before it hits the ground.0334

The acceleration in the y direction if we call down positive is 10 m/s2.0344

The only thing that is the same between these two tables is the time horizontally and the time vertically -- regardless it hits the ground at the same amount of time.0351

Whatever time we have time vertically must be the time horizontally.0360

So let us solve for the time with these things then we can solve for Δx, which is what we are after -- how far the ball travelled horizontally before reaching the ground.0364

Well to do this, if I want to find time -- Oh! We have this nice V initial y = 0.0376

So let us use delta y = V initial y x T + 1/2A yT2, and again our trick, V initial y is 0 so that whole term goes away.0382

Now then we can rearrange this to solve for t.0395

T is going to be equal to delta y/Ay -- that is T2, so we are going to need to take the square root.0399

So that is 2 x 2 m/10m/s2 square root, or 0.63 seconds.0410

If the ball is in the air 0.63 seconds vertically, it is in the air 0.63 seconds horizontally.0420

Now to find the horizontal displacement, this delta x -- delta x = Vx times t. V times t. 0427

That is 42 m/s times 0.63 seconds or 26.5 meters.0439

How far did the ball travel horizontally? 26.5 meters.0451

Let us take a look at another example problem.0460

We have a diagram here showing the path of a stunt car driven off of a cliff.0462

If we neglect friction, let us compare the horizontal component of the car's velocity at point A., where it has some horizontal component of the velocity to the horizontal component of the car's velocity at B.0467

Well the trick here is realizing that horizontally, there is no acceleration.0480

Nothing is causing the car to speed up or slow down horizontally because we are neglecting friction.0485

Therefore they have to be the same.0491

Now vertically if we looked at this we would have a smaller vertical velocity here and a bigger vertical velocity here.0495

It is speeding up in the vertical direction. Horizontally, not so much.0502

Let us take a look at some angled projectiles.0509

For objects launched at an angle you have to first break up the object's initial velocity into the x and y components of initial velocity.0512

Then you use those components to fill out your tables.0519

And again, any object will travel the maximum horizontal distance on level ground with no air resistance when you use a launch angle of 45 degrees.0522

Herman the human cannonball is launched from level ground at an angle of 30 degrees above the horizontal -- that is a great career choice, human cannonball.0535

He has an initial velocity of 26 m/s. How far does Herman travel horizontally before reuniting with the ground.0543

Well hopefully he reunites with the net right above the ground, but we won't worry about details for now.0550

Let us draw our diagram. There is our ground.0558

Herman gets launched at a velocity of 26 m/s at an angle of 30 degrees above the horizontal.0562

First thing we need to do is get the initial horizontal and vertical components of Herman's velocity.0573

Horizontally, that is going to be V initial x is 26cos30 degrees which is about 22.5 m/s.0579

The vertical component of his velocity, V initial y is going to be 26sin30.0595

Sin30 degrees is 1/2 so that is 13 m/s.0604

Now let us do our analysis.0610

If we start off by looking at the horizontal direction, let us label it.0612

We will call to the right our positive direction.0618

We have V initial x, Vx, delta x, Ax and t.0620

V initial x is 22.5 m/s. V final x has to be the same thing because Ax is 0.0628

There is no acceleration horizontally.0636

Let us try the same thing vertically.0639

Vertically, Herman is going up first, so let us call up the positive direction in this problem.0646

V initial y is equal to 13 m/s. V final y -- well, let us pull a trick in physics here.0651

Herman is going to come up and come back down.0661

As opposed to analyzing the entire range of motion, let us cut it in half again right there because it is symmetric.0663

If we can do that, then we can say you know at this highest point, his velocity in the y direction right there is 0.0671

Horizontally he still has velocity, but vertically for a split second, his velocity is 0.0679

So V final y or Vy, we can call 0. Delta y, we do not know how high he has gone yet, but we know the acceleration in the y is going to be -10 and t.0684

Let us see if we can solve for t, the time it takes for him to go up.0696

V final y is going to be V initial y + At.0702

Therefore T = Vy - V initial y/A, or that is going to be 0-13/-10.0709

Which gives us a time on the way up of 1.3 s.0722

If it takes him 1.3 s to go up, it must take him 1.3 s to go down.0728

So if the time to go up is 1.3 s, the total time in the air -- the total time that Herman is travelling horizontally must be 2.6 s.0733

Now we can find out how far he went horizontally.0744

Δx equals the velocity in the x direction times the time he is travelling.0747

Or 22.5 m/s times 2.6 s, the total time he is in the air, for a total of about 58.5 m.0752

Looking at an object launched at an angle.0768

Let us take a look at how this applies with the motion graphs we have been doing.0773

An arrow is launched from level ground with an intial velocity, V0, at some angle theta above the horizontal.0777

Let us sketch graphs of the displacement velocity and acceleration of the arrow as functions of time.0783

Well we have to do this for the x and the y direction, so let us make our graphs.0789

We have x versus x, Vx versus t, and Ax versus t.0797

All of this will be the x direction.0814

And let us do the same thing down here for the y direction.0817

Y versus time, velocity in the y direction versus time and finally, acceleration in the y versus time.0823

All right, starting with the x, we know the acceleration horizontally is 0, right. 0838

That one is pretty easy.0844

Because of that, the velocity in the x direction must be constant. It cannot change.0847

So whatever it is, it stays the same. 0852

And because it is a constant velocity we have a linear displacement or position versus time graph.0856

Trying the same thing for the y, we know the acceleration in the y is constant.0865

It could be -9.8, +9.8 m/s2, however you want to label it, I am going to draw it as if it is -9.8, but it does not change.0872

The acceleration due to gravity near the surface of the Earth are roughly constant.0880

The velocity in the y -- well, if we are calling this negative it must have started with some positive velocity on the way up.0885

Over time vertically eventually it hits that highest point, its velocity becomes 0 and it accelerates back down.0892

So we would get a graph that looks something like that.0899

Now the position in the y -- that one is pretty obvious.0902

It starts at ground level -- going to get bigger and bigger and bigger -- slow down, slow down and it is going to come back to the ground.0906

So there is our motion graphs for position time, velocity time, and acceleration time in both the x and the y positions.0914

All right, let us take a look at what happens when we launch something from a height.0926

Let us look at an example for an object that is launched from a height.0932

Blackbeard the pirate fires a cannonball from the deck of his ship at an angle of 30 degrees above the horizontal with an initial velocity of 300 m/s.0936

How far does the cannonball travel before contacting the ocean waters if the ship's deck is 8 m above the water line.0946

Well now in this case, we are launching our projectile from above the ground and it is coming back to a different vertical position.0952

We have an 8 m launch height.0968

How do we deal with that? Same thing we have been doing.0971

Let us take a look and we can say, horizontally what we are given, we know V initial in the x direction is going to be 300 m cos30 degrees, which is about 260 m/s.0975

Final velocity in the x direction has to be the same. No acceleration horizontally, so 260.0994

Delta x -- what we are trying to find -- acceleration in the x is 0 and we do not know t.1001

Vertically, let us call up our positive y direction again.1009

V initial y is going to be 300sin 30 which is 150. 1016

V final y -- we do not know. Delta y is going to be -8 and Ay is going to be -10, since we called up the positive direction.1023

Let us see what we can figure out here.1037

The first thing I am going to do is I am going to see if I cannot figure out what the final velocity in the y is.1039

That might be an interesting way as opposed to trying to go and finding t first which is going to lead us to a quadratic.1046

So Vy2 = V initial y2 + 2A delta y.1052

Therefore 1502 + 2 times -10 times-8 is going to equal Vy2, which implies then that Vy must be equal to +/-150.5.1061

Of course if we know it is going up at 150 -- +150 -- coming down that must be -150, so we are going to choose, using common sense, Vy = -150.5 m/s and now I know Vy, -150.5 m/s.1090

Finding the time then, Ay = delta Vy/T -- change in velocity over time.1112

Therefore t is going to be equal to delta Vy/Ay, or the final, -150.5 - 150/Ay, -10, or roughly, 30 s.1122

So if it is in the air for 30 s, that is the same time horizontally, we can solve for the displacement in the x position by delta x = Vt.1144

Our velocity in the x direction is 260. Our time in the x direction is 30 s = delta x.1160

So I come up with the delta x value of about 7800 m.1169

Same idea, just now in the vertical direction we do have an overall displacement.1178

If it starts here at 8 m and it ends up here at the water level, it has gone down 8 m -- -8 from where it started.1182

All right, one last problem. A nice quick one and we are done.1192

Kevin kicks a football across level ground.1196

If the ball follows the path shown, what is the direction of the ball's acceleration at point P?1199

Here at point P or any time along this path, its in projectile motion.1206

The only force acting on it, the only acceleration is the acceleration due to gravity near the surface of the Earth, 9.8 m/s2 down.1212

That is all there is to it.1226

Hope that gets you a good start on projectile motion.1227

Thanks for watching Make it a great day.1229