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Lecture Comments (8)

1 answer

Last reply by: Professor Dan Fullerton
Thu Oct 20, 2016 3:28 PM

Post by Sarmad Khokhar on October 20 at 03:06:12 PM

Your calculation in Example V11 is wrong as 260 into 30.7 is 7982 not 7970. Thanks

1 answer

Last reply by: Professor Dan Fullerton
Thu Mar 3, 2016 5:50 AM

Post by Joy Ojukwu on March 3 at 04:56:28 AM

y= 15, so why did you set it up with two y
y= 15m +.557x
my question is, y is 15m,
I taught it will be 15m = .557x +..........

1 answer

Last reply by: Professor Dan Fullerton
Thu Jan 14, 2016 7:27 AM

Post by Jerica Cui on January 13 at 11:06:37 PM

Hello Professor.
Can you explain a little bit further why an object will travel max horizontally when the launch angle is 45 degree? (why at this specific angle?)

0 answers

Post by Professor Dan Fullerton on January 5 at 06:16:52 AM

Hi Sohan.  Yes, either 9.8 or 10 is acceptable and will get you full credit.

0 answers

Post by Sohan Mugi on January 4 at 09:18:42 PM

Hello Professor Fullerton. I just had a quick question about the calculations for each of these problems. For the acceleration in the previous kinematics part 2 lecture, you have stated that it is possible to use 10m/s^2 instead of 9.8m/s^2. However, in this lecture, you seem to have used only 9.8m/s^2. I have used 10m/s^2 when I was doing these problems, and when I came to the second to last problem with the pirate cannonball, I got an answer that was about 100 meters less than the answers you have put. However, when I did the same thing again with the 9.8m/s^2, I got it to be the answer you put. Do you think that it might be better off to use the 9.8 instead of the 10 or does it makes no difference if they realize that you use 10m/s^2 in your work for the AP Physics C:Mechanics Exam FRQ and possibly still see that I did all the other calculations right and give the full credit?

Projectile Motion

  • The horizontal acceleration of a projectile in flight is zero.
  • The vertical acceleration of a projectile in flight is the acceleration due to gravity on the surface of Earth.
  • You may use the kinematic equations to separately analyze the horizontal and vertical components of a projectile’s motion.

Projectile Motion

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 0:07
  • What is a Projectile? 0:28
    • What is a Projectile?
  • Path of a Projectile 0:58
    • Path of a Projectile
  • Independence of Motion 2:45
    • Vertical & Horizontal Motion
  • Example I: Horizontal Launch 3:14
  • Example II: Parabolic Path 7:20
  • Angled Projectiles 8:01
    • Angled Projectiles
  • Example III: Human Cannonball 10:05
  • Example IV: Motion Graphs 14:39
  • Graphing Projectile Motion 19:05
    • Horizontal Equation
    • Vertical Equation
  • Example V: Arrow Fired from Tower 21:28
  • Example VI: Arrow Fired from Tower 24:10
  • Example VII: Launch from a Height 24:40
  • Example VIII: Acceleration of a Projectile 29:49

Transcription: Projectile Motion

Hello, everyone, and welcome back to www.educator.com.0000

In this lesson we are going to talk about projectile motion.0003

Our objectives for the lesson include analyzing situations of projectile motion in uniform gravitational fields.0008

Writing down expressions for both the horizontal and vertical components of velocity and positions as functions of time 0014

and using those expressions and analyzing the motion of a projectile that was projected with some arbitrary initial velocity.0020

What is a projectile?0028

A projectile is an object that is acted upon only by gravity.0030

In reality we know that air resistance plays a role but as we began our study of projectile motion we are going to neglect air resistance.0035

When we talk about projectiles typically what we are talking about is something that is launched at an angle 0042

and follows some sort of path something like that.0048

Although technically something going straight up and down as a projectile that it is not really nearly as interesting.0051

Let us look at the path of the projectile.0058

Projectiles launched an angle move in parabolic arcs.0060

Let us draw a baseline here.0066

If we have some nice flat level ground and we launch some object to some initial velocity let us call it Vi for V initial.0070

At some angle θ i with respect to the ground it is going to follow a parabolic arc.0081

It is going to come up and come back down in a symmetric path assuming we neglect air resistance.0087

A couple of interesting things to point out here.0096

However, the long it takes to go up is the exact same amount of time it takes to go down assuming it returns to level ground.0099

Its initial speed right when it leaves the ground is its speed right before it hits the ground.0107

Its initial angle is going to be the same angle it makes with the ground over here so we have this symmetry of motion.0113

We could look at the maximum height that our projectile attains which is halfway through its path.0121

There is our maximum height.0132

We could also talk about the range of our projectile.0134

How far it travels horizontally?0137

Our range would be that distance.0140

Projectiles follow parabolic paths and assuming they are coming down to the same level they left at it will follow it will have a symmetric path.0153

As we study these, one thing that will help us simplify this tremendously is looking at the independence of motion.0165

When projectiles are launched at an angle they have some piece of vertical motion and some piece of horizontal motion.0171

We can treat those 2 separately.0178

The vertical motion is acted upon by the acceleration due to gravity by the force of gravity.0180

Horizontal motion there is no acceleration because there is no air resistance.0185

As we split those up we can greatly simplify our problem solving process.0189

Let us take a look with an example.0195

Fred throws a baseball 42 m/s horizontally from a height of 2m.0197

How far will the ball travel before it reaches the ground?0203

Let us start off with a little diagram.0207

He throws it from a height of 2m.0210

It is the same as if we tossed a ball off a cliff of a height of 2m, not that much of a cliff.0213

Here is our ball and it is traveling horizontally at 42 m/s.0219

Our height here is 2m.0226

Let us take a look at what we know.0231

Horizontally we know the velocity the speed is 42 m/s and that is going to remain the same because there is no acceleration until it is acted upon by some other force.0235

Once it hits the ground at which point or into a different studying that projectile motion any longer.0245

We will have some change in horizontal position Δ x and it will take some amount of time to do this.0251

We could also look at the vertical components of its motion.0259

Vertically its initial velocity is 0.0263

It is moving horizontally but up and down vertically initially it is not moving.0268

Let us call down our positive y direction since that is the direction it moves first and the vertical.0272

We will have some final velocity we do not know that yet.0278

It will travel some vertical distance Δ y which will be 2m, +2m because we call down the positive y direction.0282

It will accelerate in the y direction at 9.8 m/s² g+ because we call down positive.0291

We do not yet know the time.0300

As we look at these what is the same between horizontal and vertical motion the amount of time the ball is in the air.0302

The time is the one thing that will be the same between our 2 tables of information.0309

If we can find the time it is in the air vertically we could then use that to figure out how far it went horizontally.0313

Let us do that.0321

Let us see if we can find a way with what we know to solve for the time vertically.0322

I think I would start with our kinematic equation Δ y = V initial t + 1/2 ay t².0328

It is nice here since V initial is 0 that whole term become 0 and we can simplify this to Δ y = 1/2 at² ay t² 0342

or t² is going to be equal to 2 Δ y / ay which will be 2 × 2m / 9.8 m/s².0356

If I want just t I take the square root of that so if I want t I will take the square root of that and find that t comes out to about 0.639s 0371

that is how long it takes for the ball to hit the ground to travel the 2m vertically.0384

That is the same as the amount of time it is traveling horizontally so that 0.639 s over there.0389

Since we have a constant velocity horizontally we can go right to our equation to find how far it travels.0397

Δ x = the velocity in the x direction × time over which it is traveling that speed or 0.639 s × velocity 42 m/s.0404

I come up with a horizontal distance that travels of 26.8m.0419

A little bit more involved but really it is just breaking up a problem into horizontal and vertical components to keep things nice and simple mathematically.0432

Let us take a look at another example.0441

The diagram represents the path of a stunt car that is driven off a cliff neglecting friction.0443

Compared to the horizontal component of the car's velocity over here at a, the horizontal component of the car's velocity at b 0449

is we need to realize here the horizontal component of velocity is not going to change because acceleration due to gravity is down.0457

It is not affecting the horizontal motion of the car.0466

Therefore, we can say the horizontal component of the velocity is the same.0469

If it asks about the vertical component of the velocity well of course it is beating up the longer it falls.0473

For objects launched at an angle in order to figure out the components of velocity we typically 0482

have to break that velocity up to x and y components using their trigonometry.0488

We will use those initial velocity components in our horizontal and vertical motion tables.0493

An object will travel the maximum horizontal distance on level ground with the launch angle of 45° as long as you are neglecting air resistance.0499

That comes up quite a bit.0507

If you want an object to go the furthest possible distance what angle do you want is 45°.0508

Let us take a look at what this would look like if we have something that is launched with a velocity at an angle.0515

There is our ground and let us say we have some initial velocity vector right there.0523

We will call that V initial.0530

It is launched at some angle θ with respect to the ground.0534

In order to find its components we will look at the x component first.0537

That piece we call the initial x and the component vertically V initial in the y.0546

From our trigonometry remember V initial y is just going to be V initial sin θ and V initial x the x component of that velocity vector 0558

is going to be V initial cos θ because we are at the adjacent side of that angle.0567

Sin θ because we are the opposite side here.0575

For range, if we want the maximum possible range on our projectile.0579

If we want to go as far as possible, there is our range.0586

The trick is a launch angle of 45° assuming it is on level ground and you are neglecting air resistance of course.0595

Let us take an example of the projectile launched in an angle.0606

Herman, the human cannonball is launched from level ground at an angle of 30° above the horizontal with the initial velocity of 26 m/s.0610

Interesting career path for Herman.0618

How far does he travel horizontally before reuniting with the ground?0621

Now let us start by taking a look at his initial velocity vector.0626

I am going to make some axis here.0633

These initially launched an angle of 30° above the ground, above the horizontal and that initial velocity is 26 m/s.0638

The x component the horizontal component of that velocity we can find the initial x is just going to be 26 m/s cos 30° or 22.5 m/s.0654

The y component the initial y will be 26 m/s sin 30° or about 13 exactly 13 m/s sin 30 is ½.0677

How far did he travel horizontally before reuniting with the ground?0699

Again, I think the most important thing for us to know here is how long he is in the air.0703

We will start by answering that with a vertical analysis.0707

Vertically we have V initial, V final, some horizontal displacement, ay and t.0712

Herman's path is going to take him up and down like that.0722

Remember our trick for simplifying this is let us look at either the way up or the way down that by cutting this in half.0726

Let us say that we want to look at the way up.0732

Time to go up, time to go down, and V initial then if we are going up we are going to call positive y up.0740

V initial is 13 m/s because we said up was positive, V final at its highest point it stops.0751

We do not know the Δ y.0759

We know the acceleration is -9.8 m/s² negative because we call up positive and the acceleration is down due to gravity.0760

We are trying to find the time and we can use V = V initial + at.0769

We arrange that to find the time is V - V initial / a is 0 -13 m/s / -9.8 m/s².0780

Our time is going to be about 1.33s but note that is the time just to go up.0794

If it takes 1.33s to go up how long does it take to go up and down?0800

Twice that so our total time is twice the time it takes to go up or 2.6 and if we do not round here I get closer to 2.65s.0806

We know the time in the air and we can figure out the range how far Herman travels horizontally.0820

Let us do that.0826

Horizontally we do not know Δ x yet but we know the velocity in the x direction was our 22.5 m/s.0828

We found that out initially by breaking at vector in the components.0841

And the total time Herman's traveling horizontally is 2.65s.0845

We just found that by analyzing the vertical motion.0850

Δ x = the velocity in the x × the time which is 22.5 m/s × 2.65s for 59.6m.0857

Let us take a look at some motion graphs then.0877

An arrow was launched from level ground with initial velocity V0 at an angle of θ above the horizontal.0881

Sketch a graph of the displacement velocity and acceleration of the arrow was functions of time and neglect air resistance.0887

Let us do this for both x and the y so we are going to make some graphs here.0894

Make another graph for velocity and another graph over here for our acceleration.0905

This those a little bit but we will get the idea across I think and our x axis.0917

We will have a look at the x up here and look at the y down here.0944

Take a look at graphs of position vs. time, velocity in the x vs. time.0948

Acceleration in the x vs. Time and for the y we will have Δ y for time and we will have the velocity 0957

and the y compared the time and will have acceleration in the y in time.0965

Let us start off on the right hand side with acceleration usually those are an easy place to start.0973

In the x direction we know there is no acceleration so that graph is nice and flat at 0 no acceleration.0979

In the y direction the entire time our acceleration is 9.8 m/s² down.0988

Assuming we are calling up positive let us just draw our line down here at -9.8 m/s² for acceleration.0995

Now let us move backwards and take a look at the velocity.1004

In the x direction the velocity must be constant.1008

We have some constant velocity and that makes sense because if we take the slope of velocity which we will get our acceleration.1012

The slope of a straight line is 0 so our acceleration is 0.1020

Down here for the y we have a negative acceleration which means our velocity must have a negative slope.1025

For something launched from level ground with initial velocity of V 0 at an angle it is going to start off that is maximum velocity.1031

It is going to slow down stop and then start going faster and faster in the negative direction.1039

There is our graph of velocity vs. time in the y direction.1050

Note that the slope is negative in constant and our acceleration is negative in constant.1054

Now for the displacement vs. time.1060

We know that we are changing our position horizontally at a constant rate.1063

We would see something that looks kind of like this.1070

Assuming we are starting from the point we call 0.1072

We are going to have a 0 nice straight linear line slightly redundant.1075

The slope of this is positive in constant.1082

I will look positive in constant velocity.1086

All of these match up when we look at our slopes.1088

When we look at displacement in the y direction what we are going to have change in values here 1092

because we have a constant but we have a change in value here so we have changing slopes.1098

Right in the middle the value of our velocity is 0, right in the middle we must have a slope of 0.1103

I would draw something kind of like this for our vertical displacement.1108

We start off and 0 we reach a maximum here halfway through the slope here is 0 corresponding to a velocity of 0. 1117

Then we start coming back the other way steeper and steeper negative slope more and more negative velocity.1125

That is how I would graph our displacement velocity and acceleration for each of these components for an object that is launched from the ground.1132

Graphing projectile motion.1145

Contract a vector component as a function of time or plot their path as a y = function of x.1147

We have been doing is function of time but if we wanted to do this is as a function of x solve for the time in the horizontal 1152

and then we can use that to eliminate time in the vertical equation.1159

What does that mean?1163

Let us look for an analysis of something launched an angle horizontally Δ x is our initial x velocity × time.1164

And therefore time would be our horizontal displacement divided by our horizontal velocity.1177

When we move to our vertical analysis Δ y = Vy initial t + 1/2 ay t².1187

We can now substitute in what we know for time from our horizontal equation recognizing that t = Δ x / our initial x velocity.1200

Δ y = V initial y × our time Δ x / V x initial + ½ ay × Δ x² / x initial².1213

Or just cleaning this up a little bit we could write this as our y position.1237

Δ y assuming we are starting at 0 equals if we do this as a function of x we are going to have Vy initial / Vx initial × 1245

our x coordinate + we will have ay / 2 Vx initial² × x².1258

Y = some constant × x + another constant x² that is the form of a parabola, a parabolic motion.1274

Let us take a look at an example here.1288

We got an air fighter from a tower, if an arrow was fired from a 15m high tower with an initial velocity 50 m/s and angle of 30° above the horizontal, 1290

find the height of the projectile above the ground as a function of its horizontal displacement.1300

Since our object goes up first let us call up our positive y direction.1307

We can take a look and figure out our initial x velocity, Vx initial is going to be V cos θ, V cos 30° or 50 m/s cos 30° which will be 43.3 m/s.1312

Doing the same thing for the y the initial velocity in the y will be V sin θ, V sin 30° or 50 sin 30 sig 30 is half so that is just going to be 25 m/s.1330

Since we called up the positive direction our acceleration in the y is -g or -9.8 m/s².1346

We can go back to our form for the motion of a projectile in parabolic form.1355

We had y = V initial / Vx initial x + ay / 2 Vx initial² x².1363

Substituting in our values that we just determined here y = 25 / 43.3 x + -9.8 / 2 × 43.3² x².1379

Let us set our origin to 0, 15 because that is where we are starting.1403

I get that y = our 15m we are starting with + 25 / 43.3, .577x - 9.8 / the quantity of 2 × √43.3 it is going to be 0.00261 x².1415

And there is our formula for the parabola that shows our x and y motion.1438

Let us actually check this and let us make a graph of that.1444

I plotted a nice pretty one in a software package that shows this formula y is a function of x.1448

Notice that we start here at 15 m are going up, coming back down and we are going to hit the ground right around there.1456

It should be pretty reasonable to say 1234 looks like we are about 244 m or so that is how far 1465

that arrow would travel before it struck the ground when launch from that 50m high tower.1472

Alright speaking of launching things from a height.1480

Black Bear the pirate fires a cannonball from the deck of his ship an angle of 30° above the horizontal with an initial velocity of 300 m/s.1483

How far does the cannonball travel before contacting the ocean waters if the ship's deck is 8m above the waterline?1494

A quick diagram here if there is our ground we are starting from 8m above and doing something like that.1503

We have got 8m as our initial height.1512

We will let us say let us take a look at our horizontal analysis.1519

Horizontally our initial velocity in the x direction is going to be 300 cos 30 or about 260 m/s.1524

Our final velocity is the same.1537

There is no acceleration horizontally so that is 260 m/s.1539

Our acceleration in the x is 0.1543

We do not know how far it goes horizontally that would be nice to know, that is what we are after.1546

We do not know time yet.1550

How are we going to figure out time?1553

Now I would go to the vertical.1556

Let us take a look at our vertical analysis to figure out some more information.1559

Vertically let us call up the positive direction since a cannonball travels up vertically first.1564

V initial in the y direction is going to be 300 m/s sin 30° or 150 m/s.1572

Let us say we have the final y we do not know yet.1584

Δ y while the entire trip if it starts here and it ends up 8m below its total displacement is going to be -8m.1592

-8 because we call up positive it is going to travel much further vertical distance but its displacement will be -8.1601

Acceleration in the y will be -9.8 m/s² negative because we called up the positive direction.1609

Let us see what we can find out here.1618

As I do this it looks like we could solve for Vy first if we try to find t initially we are going to come up with a quadratic.1621

Let us go with the finding final velocity in the y first.1631

Vy² = V initial y² + 2ay Δ y our kinematic equation complies that Vy² is going to be equal to our initial velocity 1635

in the vertical 150² + 2 × -9.8 × -8.1648

Vy² = 22,657 m² / s² or if I take the square root of that Vy = + or - about 150.5 m/s.1658

We are going to take some common sense here.1676

Go here when it gets this point is it traveling up or down.1679

It must be going down vertically because we called up positive so we must be negative value Vy is -150.5 m/s.1683

Find the velocity in the y -150.5 m/s.1700

From here we can figure out the time it took to do that.1707

Using another kinematic equation let us go with if ay is Δ Vy / t then t is going to be Δ Vy / a.1712

That is going to be our -150.5 m/s -150 m/s our initial velocity divided by acceleration -9.8 m/s² or about 30.7s.1725

T= 30.7s and that is the same amount of time that the cannonball is going to be traveling horizontally.1743

I can fill in my time here horizontally as 30.7s.1751

We can take a look now at the horizontal displacement of the cannonball.1757

Δ x is going to be the x velocity × time or 260 m/s the horizontal velocity our time of 30.7s or about 7,970 m.1762

It is quite the path that is traveling.1784

Taking a look at one last question here.1790

Kevin kicks a football across level ground, if the ball follows the path shown what is the direction of the balls acceleration when it gets to point b?1793

Fairly common question but a trick question.1802

The entire time the ball is in the air with the only force acting on it is gravity.1806

Gravity pulls down so the acceleration on the ball is a very pretty down, straight down there is the direction of our acceleration.1812

Any point in the path even when vertically it stops at the highest point the acceleration is always straight down.1824

Thank you for watching www.educator.com, we will see you again soon.1830

Make it a great day everyone.1833