1. 3
Motion in a Plane

2. Ch. 3: Projectile Motion
2 D Kinematics

3. Projectile Motion
A curvilinear motion influenced by gravity only is called projectile motion. The path of this motion is always going to be an arc of a parabola
Note: A parabola is the graphical representation of a quadratic equation.
To study this motion we will use the technique that was presented earlier in this module which consists in breaking down (resolving) a curvilinear motion into two independent linear motions.

4. Horizontal Projection
Let's consider first the simpler case of an object that is launched with a horizontal velocity vo from a certain height. For example, in the balcony scene, Juliet throws horizontally the key of the entrance door towards Romeo, but also by negligence simultaneously knocks of a pot of flowers. vo
Both objects will hit the ground at the same time, and here is why:
The motion of the key can be broken down into a linear uniform horizontal motion with a constant velocity vo, and a vertical motion with no downward initial velocity but with the acceleration "g" (free fall). If we consider the origin of the motion in the balcony, we can write the following equations to describe the distances traveled by the key in the two directions:
Horizontally: x = vot (1)
Vertically : y = ½gt (2) x y

5. Horizontal Projection Example
If the height of the balcony above the ground is let's say 4.9 m, and we substitute this value in equation (2) we get:
4.9 m = ½ x (9.8 m/s2) x t2
The solution of this simple equation is t = 1 s. This represents the time elapsed from the moment the key was thrown to the moment the key reached the ground.
The motion of the pot is a simple free fall described by an equation identical to (2) which evidently would give the same time of fall t = 1 s.

6. Velocity
Note: The horizontal velocity stays the same throughout the motion of the projectile.
In the diagram above, both objects are at the same height at any time. They will also have the same vertical velocity:
vy = gt (3)
represented by the downward arrows which will have the same length for both objects at any time.

7. Velocity in Horizontal Projectile Motion
An object projected horizontally has an initial velocity in the horizontal direction, and acceleration (due to gravity) in the vertical direction. The time it takes to reach the ground is the same as if it were simply dropped.

8. Examples
Shoot an arrow horizontally with vo = 20 m/s, from a height of h = 2m.
A) How long is the arrow in the air?
B) What is the vertical velocity when it lands?
C) How far will it land?
h = gt2/ t = √(2h/g) = √(2 x 2m/9.8 m/s2 )= 0.64 s
vy = gt = 9.8 x 0.64 = 6.26 m/s
R = vot = 20 m/s x .64 s = 12.8 m

9. Projection at an Arbitrary Angle
This time we will assume that the projectile can be launched at any angle above the horizontal. We will study this situation by applying the method of resolving the curvilinear motion into two independent linear motions along the horizontal and vertical directions, exactly as we did in the previous case.
A good example of this type of projectile motion would be the motion of a ball kicked by a soccer player with an initial velocity of vo = 10 m/s at an angle of 30o above the horizontal.

10. Projectile Velocity Components
The components of the initial velocity vo are:
Horizontal:
vox = vocos (4)
Vertical:
voy = vosin (5)
vv vo -g
 vh
Launching and same elevation.

11. Projectile Motion
A projectile launched in an arbitrary direction may have initial velocity components in both the horizontal and vertical directions, but its acceleration is still downward.

12. Equations of Motion
If we neglect the air resistance, there is nothing that would affect the horizontal component of the velocity. Therefore the horizontal component vh stays constant and the horizontal motion is linear uniform. The equation of motion for this direction is:
x = vox t = (vocos)t (6)
The vertical motion is an "upward" free fall with the acceleration -g. The vertical velocity is:
vy = voy – gt = vosin - gt (7)
The height above the ground at any time is given by the expression:
y = voyt – ½gt2= (vosin)t – ½gt2 (8)
Note: Equations (7) and (8) are similar to the equations…. from the “Free Fall” section, except vo is replaced by voy, since in this case only the vertical component of the velocity is causing the upward motion.

13. Range
A very important characteristic of the projectile motion is the total horizontal distance traveled by the projectile, called range denoted by R. To find the range we will first compute the time spent by the projectile in the air. The time tup for the projectile to reach the maximum height can be found from equation (7) putting vy = 0, because at the “apex” of his path the projectile ceases to ascend. Therefore:
0 = voy – gtup = vosin - gtup
Solving for tup we get:
tup = (vosin)/g (9)

14. Range
Because of the symmetry of this motion, the time tdown spent by the projectile on the descending part of the motion must be equal to tup. We can write:
ttot = tup + tdown = 2 tup = 2(vosin)/g (10)
Finally, to get the range R we will substitute ttot in equation (6), since ttot represents also the time spent in horizontal motion.
R = (vocos) ttot = (vocos) x 2(vosin)/g
Rearranging we obtain:
R = (vo2/g) x (2 sin cos)
Using a double angle identity from trigonometry sin2 = 2 sin cos we get the final version of this equation:
R = (vo2/g)sin2 (11)
For a given initial velocity vo the range depends solely on the angle .
If  = 45o, sin (2 x 45o) = sin 90o = 1 which is the maximum value of the sine function and implicitly the maximum value of R denoted Rmax. Its value is:
Rmax = vo2/g

15. Range
To get a better picture on how the range R depends upon the angle “”, we graphed the path of a projectile launched with vo = 10 m/s at different angles:
Notice that the maximum range is obtained for a launch at 450.
Is also interesting to observe that for complementary pairs of angles like for example 150 and or and 600 , the range is the same. This is due to a property of the sine function which gives the same value for suplementary angles (we have 2θ in the formula and “2 x complementary = suplementary”, so to speak).

16. Paths in Projectile Motion
The range of a projectile is maximum (if there is no air resistance) for a launch angle of 45°.

17. Projectile Motion
With air resistance, the range is shortened, and the maximum range occurs at an angle less than 45°.

18. Example 1 vy = 0, y = ymax voy = vosinθ
Arrow shot at θ = 30o, spends ttot = 3s in the air.
a) What is the initial velocity?
tup = 1.5s
b) What is the max height?
vy = 0, y = ymax
voy = vosinθ
c) What is the range?

19. Additional question: How many bones did the biker break?
Example2
A biker performing a stunt is riding of the flat roof of a building and R=4m from the base of the wall. What is the horizontal velocity of the biker when he left the building if he spends t=1s in the air?
If t = 1 s, x = R
x = vot = R
vo x 1s = 4 m, vo = 4m/s vo
Additional question: How many bones did the biker break?
<206 4m

20. Example 3 ttot = 2 x 1.25s = 2.5 s Know vo = 25 m/s, θ = 30o.
Find ttot, range, ymax
ttot = 2 x 1.25s = 2.5 s

21. Projectile Motion Review
Horizontal Projection:
Horizontally: x = vot (1)
Vertically : y = ½ gt (2)
Velocity: vy = gt (3)
an arbitrary angle: launching same elevation as landing
The components of the initial velocity vo are:
Horizontal: vox = vocos (4)
Vertical: voy = vosin (5)

22. Launching = Landing Review
Horizontal motion (OX)is linear uniform. x = vox t = (vocos)t (6)
The vertical motion is an "upward" free fall with the acceleration -g. The vertical velocity is vy = voy – gt = vosin - gt (7)
or:
The height above the ground (OY):
y = voyt – ½ gt2 = (vosin)t – ½ gt2 (8)
Other important formulas: (9) time up, max. height(10), Range (11)

23. Equations of Motion General
The equation of motion in the Ox direction is:
x =xo+ vox t =x0 + (vocos)t (12)
The vertical velocity is:
vy = voy – gt = vosin - gt (13)
The vertical component on Oy:
y = y0 +voyt – ½ gt2= yo+(vosin)t – ½ gt2 (14) y vo
xo, yo x

24. Soccer Example y = y0 +voyt – gt2/2 = yo+(vosin)t – gt2/2 x = vox t
4) A soccer ball is kicked from the edge of a 25.0 m high cliff. The ball leaves the cliff with a velocity of 15.2 m/s at an angle of 18.0o above the horizontal.
18o
15.2 m/s
What is the horizontal component of the ball’s initial velocity?
What is the vertical component of the ball’s initial velocity?
How far does the ball travel horizontally before hitting the ground?
a) vox = vocos = 15.2cos18 = m/s
b) voy = vosin = 15.2 sin18 = 4.70 m/s
y = y0 +voyt – gt2/2 = yo+(vosin)t – gt2/2
x = vox t
X = x 2.79 = 40.3 m
0 = t- 9.8t2/2
t=2.79 s

25. ConcepTest 3.5 Dropping a Package
1) quickly lag behind the plane while falling
2) remain vertically under the plane while falling
3) move ahead of the plane while falling
4) not fall at all
You drop a package from a plane flying at constant speed in a straight line. Without air resistance, the package will:
[CORRECT 5 ANSWER]

26. ConcepTest 3.5 Dropping a Package
1) quickly lag behind the plane while falling
2) remain vertically under the plane while falling
3) move ahead of the plane while falling
4) not fall at all
You drop a package from a plane flying at constant speed in a straight line. Without air resistance, the package will:
Both the plane and the package have the same horizontal velocity at the moment of release. They will maintain this velocity in the x-direction, so they stay aligned.
[CORRECT 5 ANSWER]
Follow-up: What would happen if air resistance were present?

27. ConcepTest 3.6a Dropping the Ball I
1) the “dropped” ball
2) the “fired” ball
3) they both hit at the same time
4) it depends on how hard the ball was fired
5) it depends on the initial height
From the same height (and at the same time), one ball is dropped and another ball is fired horizontally. Which one will hit the ground first?
[CORRECT 5 ANSWER]

28. ConcepTest 3.6a Dropping the Ball I
1) the “dropped” ball
2) the “fired” ball
3) they both hit at the same time
4) it depends on how hard the ball was fired
5) it depends on the initial height
From the same height (and at the same time), one ball is dropped and another ball is fired horizontally. Which one will hit the ground first?
Both of the balls are falling vertically under the influence of gravity. They both fall from the same height. Therefore, they will hit the ground at the same time. The fact that one is moving horizontally is irrelevant – remember that the x and y motions are completely independent !!
[CORRECT 5 ANSWER]
Follow-up: Is that also true if there is air resistance?

29. ConcepTest 3.6b Dropping the Ball II
1) the “dropped” ball
2) the “fired” ball
3) neither – they both have the same velocity on impact
4) it depends on how hard the ball was thrown
In the previous problem, which ball has the greater velocity at ground level?
[CORRECT 5 ANSWER]

30. ConcepTest 3.6b Dropping the Ball II
1) the “dropped” ball
2) the “fired” ball
3) neither – they both have the same velocity on impact
4) it depends on how hard the ball was thrown
In the previous problem, which ball has the greater velocity at ground level?
Both balls have the same vertical velocity when they hit the ground (since they are both acted on by gravity for the same time). However, the “fired” ball also has a horizontal velocity. When you add the two components vectorially, the “fired” ball has a larger net velocity when it hits the ground.
[CORRECT 5 ANSWER]
Follow-up: What would you have to do to have them both reach the same final velocity at ground level?

31. ConcepTest 3.6c Dropping the Ball III
1) just after it is launched
2) at the highest point in its flight
3) just before it hits the ground
4) halfway between the ground and the highest point
5) speed is always constant
A projectile is launched from the ground at an angle of 30o. At what point in its trajectory does this projectile have the least speed?
[CORRECT 5 ANSWER]

32. ConcepTest 3.6c Dropping the Ball III
1) just after it is launched
2) at the highest point in its flight
3) just before it hits the ground
4) halfway between the ground and the highest point
5) speed is always constant
A projectile is launched from the ground at an angle of 30o. At what point in its trajectory does this projectile have the least speed?
The speed is smallest at the highest point of its flight path because the y-component of the velocity is zero.
[CORRECT 5 ANSWER]

33. ConcepTest 3.7a Punts I
Which of the 3 punts has the longest hang time?
4) all have the same hang time 1 2 3 h
[CORRECT 5 ANSWER]

34. ConcepTest 3.7a Punts I
Which of the 3 punts has the longest hang time?
4) all have the same hang time 1 2 3 h
The time in the air is determined by the vertical motion ! Since all of the punts reach the same height, they all stay in the air for the same time.
[CORRECT 5 ANSWER]
Follow-up: Which one had the greater initial velocity?

35. ConcepTest 3.7b Punts II 1 2 3) both at the same time
A battleship simultaneously fires two shells at two enemy submarines. The shells are launched with the same initial velocity. If the shells follow the trajectories shown, which submarine gets hit first ? 1 2
[CORRECT 5 ANSWER]
3) both at the same time

36. ConcepTest 3.7b Punts II 1 2 3) both at the same time
A battleship simultaneously fires two shells at two enemy submarines. The shells are launched with the same initial velocity. If the shells follow the trajectories shown, which submarine gets hit first ?
The flight time is fixed by the motion in the y-direction. The higher an object goes, the longer it stays in flight. The shell hitting submarine #2 goes less high, therefore it stays in flight for less time than the other shell. Thus, submarine #2 is hit first. 1 2
[CORRECT 5 ANSWER]
3) both at the same time
Follow-up: Which one traveled the greater distance?

37. ConcepTest 3.8 Cannon on the Moon
For a cannon on Earth, the cannonball would follow path 2. Instead, if the same cannon were on the Moon, where g = 1.6 m/s2, which path would the cannonball take in the same situation? 1 2 4 3

38. ConcepTest 3.8 Cannon on the Moon
For a cannon on Earth, the cannonball would follow path 2. Instead, if the same cannon were on the Moon, where g = 1.6 m/s2, which path would the cannonball take in the same situation?
The ball will spend more time in flight because gMoon < gEarth. With more time, it can travel farther in the horizontal direction. 1 2 4 3
Follow-up: Which path would it take in outer space?

39. ConcepTest 3.9 Spring-Loaded Gun
The spring-loaded gun can launch projectiles at different angles with the same launch speed. At what angle should the projectile be launched in order to travel the greatest distance before landing?
1) 15°
2) 30°
3) 45°
4) 60°
5) 75°

40. ConcepTest 3.9 Spring-Loaded Gun
The spring-loaded gun can launch projectiles at different angles with the same launch speed. At what angle should the projectile be launched in order to travel the greatest distance before landing?
1) 15°
2) 30°
3) 45°
4) 60°
5) 75°
A steeper angle lets the projectile stay in the air longer, but it does not travel so far because it has a small x-component of velocity. On the other hand, a shallow angle gives a large x-velocity, but the projectile is not in the air for very long. The compromise comes at 45°, although this result is best seen in a calculation of the “range formula” as shown in the textbook.

41. ConcepTest 3.10a Shoot the Monkey I
You are trying to hit a friend with a water balloon. He is sitting in the window of his dorm room directly across the street. You aim straight at him and shoot. Just when you shoot, he falls out of the window! Does the water balloon hit him?
1) yes, it hits
2) maybe–it depends on the speed of the shot
3) no, it misses
4) the shot is impossible
5) not really sure
[CORRECT 5 ANSWER]
Assume that the shot does have enough speed to reach the dorm across the street.

42. ConcepTest 3.10a Shoot the Monkey I
You are trying to hit a friend with a water balloon. He is sitting in the window of his dorm room directly across the street. You aim straight at him and shoot. Just when you shoot, he falls out of the window! Does the water balloon hit him?
1) yes, it hits
2) maybe–it depends on the speed of the shot
3) no, it misses
4) the shot is impossible
5) not really sure
Your friend falls under the influence of gravity, just like the water balloon. Thus, they are both undergoing free fall in the y-direction. Since the slingshot was accurately aimed at the right height, the water balloon will fall exactly as your friend does, and it will hit him!!
[CORRECT 5 ANSWER]
Assume that the shot does have enough speed to reach the dorm across the street.

43. ConcepTest 3.10b Shoot the Monkey II
You’re on the street, trying to hit a friend with a water balloon. He sits in his dorm room window above your position. You aim straight at him and shoot. Just when you shoot, he falls out of the window! Does the water balloon hit him??
1) yes, it hits
2) maybe–it depends on the speed of the shot
3) the shot is impossible
4) no, it misses
5) not really sure
[CORRECT 5 ANSWER]
Assume that the shot does have enough speed to reach the dorm across the street.

44. ConcepTest 3.10b Shoot the Monkey II
You’re on the street, trying to hit a friend with a water balloon. He sits in his dorm room window above your position. You aim straight at him and shoot. Just when you shoot, he falls out of the window! Does the water balloon hit him??
1) yes, it hits
2) maybe–it depends on the speed of the shot
3) the shot is impossible
4) no, it misses
5) not really sure
This is really the same situation as before!! The only change is that the initial velocity of the water balloon now has a y-component as well. But both your friend and the water balloon still fall with the same acceleration -- g !!
[CORRECT 5 ANSWER]
Assume that the shot does have enough speed to reach the dorm across the street.

45. ConcepTest 3.10c Shoot the Monkey III
You’re on the street, trying to hit a friend with a water balloon. He sits in his dorm room window above your position and is aiming at you with HIS water balloon! You aim straight at him and shoot and he does the same in the same instant. Do the water balloons hit each other?
1) yes, they hit
2) maybe–it depends on the speeds of the shots
3) the shots are impossible
4) no, they miss
5) not really sure

46. ConcepTest 3.10c Shoot the Monkey III
You’re on the street, trying to hit a friend with a water balloon. He sits in his dorm room window above your position and is aiming at you with HIS water balloon! You aim straight at him and shoot and he does the same in the same instant. Do the water balloons hit each other?
1) yes, they hit
2) maybe–it depends on the speeds of the shots
3) the shots are impossible
4) no, they miss
5) not really sure
This is still the same situation!! Both water balloons are aimed straight at each other and both still fall with the same acceleration -- g !!
Follow-up: When would they NOT hit each other?