For more information, please see full course syllabus of AP Physics C: Mechanics

For more information, please see full course syllabus of AP Physics C: Mechanics

## Discussion

## Study Guides

## Download Lecture Slides

## Table of Contents

## Transcription

## Related Books

### 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
- Objectives
- What is a Projectile?
- Path of a Projectile
- Independence of Motion
- Example I: Horizontal Launch
- Example II: Parabolic Path
- Angled Projectiles
- Example III: Human Cannonball
- Example IV: Motion Graphs
- Graphing Projectile Motion
- Example V: Arrow Fired from Tower
- Example VI: Arrow Fired from Tower
- Example VII: Launch from a Height
- Example VIII: Acceleration of a Projectile

- 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

### AP Physics C: Mechanics Online Course

I. Introduction | ||
---|---|---|

What is Physics? | 7:12 | |

Math Review | 1:00:51 | |

II. Kinematics | ||

Describing Motion I | 23:47 | |

Describing Motion II | 36:47 | |

Projectile Motion | 30:34 | |

Circular & Relative Motion | 30:24 | |

III. Dynamics | ||

Newton's First Law & Free Body Diagrams | 23:57 | |

Newton's Second & Third Laws of Motion | 23:57 | |

Friction | 20:41 | |

Retarding & Drag Forces | 32:10 | |

Ramps & Inclines | 20:31 | |

Atwood Machines | 24:58 | |

IV. Work, Energy, & Power | ||

Work | 37:34 | |

Energy & Conservative Forces | 28:04 | |

Conservation of Energy | 54:56 | |

Power | 16:44 | |

V. Momentum | ||

Momentum & Impulse | 13:09 | |

Conservation of Linear Momentum | 46:30 | |

Center of Mass | 28:26 | |

VI. Uniform Circular Motion | ||

Uniform Circular Motion | 21:36 | |

VII. Rotational Motion | ||

Rotational Kinematics | 32:52 | |

Moment of Inertia | 24:00 | |

Torque | 26:09 | |

Rotational Dynamics | 56:58 | |

Angular Momentum | 33:02 | |

VIII. Oscillations | ||

Oscillations | 1:01:12 | |

IX. Gravity & Orbits | ||

Gravity & Orbits | 34:59 | |

X. Sample AP Exam | ||

1998 AP Practice Exam: Multiple Choice | 28:11 | |

1998 AP Practice Exam: Free Response Questions (FRQ) | 28:11 |

### 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

1 answer

Last reply by: Professor Dan Fullerton

Thu Oct 20, 2016 3:28 PM

Post by Sarmad Khokhar on October 20, 2016

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, 2016

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, 2016

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, 2016

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, 2016

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?