1. PROJECTILE MOTION
Chapter 3, sections , 3.5

2. EXAMPLES OF PROJECTILE MOTION

3. PROJECTILE MOTION
vx: Horizontal Velocity. Once in air, vx does not change since there is no force pushing or pulling horizontally. vx v1 v2
vy: Vertical Velocity. Once in air, vy changes because of the force of gravity. The projectile accelerates at the rate of 9.8 m/s2 downward. v3
PM simulation

4. PROJECTILE MOTION
A projectile is an object shot through the air upon which the only force acting is gravity (neglecting air resistance).
The trajectory (path through space) or a projectile is parabolic.

5. (for horizontally launched projectile):
Time of flight
(for horizontally launched projectile):
Time it takes to hit the ground is completely unaffected by the horizontal velocity, only depends on height

6. Horizontal and vertical parts of the motion are independent
The horizontal part of a projectile motion problem is a constant velocity problem.
The vertical part of a projectile motion problem is a free fall problem.
Horizontal and vertical parts of the motion are independent
To solve a projectile motion problem, break up the motion into x and y components.

7. Kinematics Equations for uniformly accelerated motion
Horizontal (ax=0)
Vertical (FREE FALL)

8. Some Common Projectile Motion Problems
Physics C
Horizontal Launch v0
When object is launched horizontally:
v0y = 0
Dx, Range
Angle Launch on Level Ground
When object returns to the same height from which it was launched:
Dy = 0 v0
Dx, Range
Bertrand

9. Remember… To work projectile problems…
Physics C
Remember… To work projectile problems…
…resolve the initial, launch velocity into horizontal and vertical components.
v0y = vo sin  Vo 
vx = vo cos 
Bertrand

10. Projectile Launched from Any Angle
At max height, vy = 0 vx vt vx
a = constant
along y vx
vty vx vx v0
Dymax = height
v0y q vx vx
v = constant along x x
Dxmax = range vf
Initial Horizontal velocity
vy = -v0y at same height landing
Initial Vertical velocity
(determines air time)
PM simulation 10

11. EXAMPLE x y t = 1.6 s 12.5 m v0 = 41.9 m/s v = 44.7 m/s, 210 below the
A golfer is on the edge of a 12.5 m bluff overlooking the 18th hole which is located 67.1 m from the base of the bluff. She launches a horizontal shot that lands in the hole on the fly. The gallery erupts in cheers.
How long was the ball in the air?
What was the ball’s initial velocity? c) What was the ball’s impact velocity (velocity right before landing) x y
t = 1.6 s
v0 = 41.9 m/s
v = 44.7 m/s, 210 below the
horizontal
12.5 m
67.1 m

12. EXAMPLE
In the circus, a clown is launched from a cannon at 40 m/s, 60o from the horizontal. Where should the other clowns hold the net so that the projectile clown lands unharmed (at the same level)?
t = 7.07 s
Dx = 141 m
v0 = 40 m/s
v0y vx

13. Where does the ball land? How far has the truck traveled in that time?
EXAMPLE
Where does the ball land? How far has the truck traveled in that time?
Ball x y
v = 10 m/s
v = 30 m/s
t = 2.0 s
Dx = 60 m B1 B2

14. EXAMPLE
If Agent Tim is flying at a constant velocity of 86.0 m/s horizontally in a low flying helicopter at an altitude of 125 m. His mission is to drop a 50 kg explosive onto a master criminal’s car. How far before the car should he release the 50 kg bomb so that it lands on the car? (ignore air resistance)
86 m/s
125 m
Bomb
Dx = ?
430m

15. Cargo Drop

16. EXAMPLE A cannon ball is fired out of a cannon at muzzle velocity of 30 m/s, 40o above the horizontal. The cannonball is fired from a m high cliff. How far from the base of the cliff will the cannonball land? What is its final velocity, right before landing?
v0 = 30 x y
40o
= m/s
= 6.90s
100.0 m
100.0 m
= 6.90s
Dx = m vx
v = 53.2m/s,
65o below the
horizontal Dx vy

17. EXAMPLE A professional soccer player is 25
EXAMPLE A professional soccer player is 25.8 m from the goal and kicks a hard shot at ground level. The ball hits the cross bar on its way down, 2.44 m above the ground, 1.98 s after being kicked. What was the initial velocity of the ball (find the x- and y- components)? x y
vx = 13.0 m/s
v0y = m/s
v0 = 17.1 m/s, 40o
2.44 m
25.8 m

18. EXAMPLE
What horizontal firing (muzzle) velocity splashes the cannonball into the pond?
x-dir
y-dir v0
t = 2.86 s
v0 = 33.2 m/s
40 m
95 m

19. Great simulations of Projectile Motion
Any type of projectile (by Mr Walsh)
Projectile over level ground
Exploring Concepts: Several projectiles with
same range OR
Same max height OR
Same initial speed

20. For the case of projectiles launched on level ground, you found that the initial conditions of launch speed, v0, and launch angle, q,
affect the projectile’s flight time (t), maximum height (Dymax), and horizontal range, (Dx) in the following ways.
Time of flight increases with v0y, the vertical component of v0
Max height increases with v0y, vertical component of v0
Horizontal Range affected by both components of v0 so that if v0 stays the same, max range is at 45o and 2 complimentary launch angles at the same speed gave the same range
For a projectile launched over level ground
DERIVE EXPRESSIONS for
time of flight (t),
max height (Dymax), and
horizontal range (Dx)
in terms of v0, q, and g.
Confirm that the derived equations
predict the observations above.
For the range equation Use the trig identity : 2sinqcosq = sin(2q)
All launched at same speed, v0

21. Time of flight Max height Range
Projectile Launched from Any Angle Over Level Ground
Shown are all launched at same speed, v0
simulation
exploring concepts simulation
When starting and landing heights are the same:
Time of flight
All depend on angle and initial velocity
Max height
Range
Max at 450

22. Projectile Launched from Any Angle Over Level Ground
Time of flight
Max height
Range

23. Rank time in air from greatest to least:
Red = green = blue
If the time in air for all was 6 sec
What is the initial vertical velocity
What is the max height?
30 m/s
45m (vavt)

24. For a Horizontally Launched Projectile a) Derive an expression for the flight time, t, and horizontal range, Dx, of a horizontally launched projectile in terms of the initial conditions: v0, h and g.
b) If the projectile was launched horizontally from double the height, what would happen to the flight time and range? v0
Time of Flight
Range

25. DO NOW
Sketch the following motion graphs for a horizontally launched projectile. x y t t vx vy t t ax ay t t 25

26. Rank flight time from greatest to least:
B=E > A=D=F > C v0 2h m v0 h m
½v0 ½h m
(A)
(B)
(C)
2v0 2h 2m v0 h 2m
2v0 h m
(D)
(E)
(F)

27. Rank range from greatest to least:
E > B> A > D > C v0 h m
2v0 h m
½v0 ½h m
(A)
(B)
(C)
2v0 2h 2m
½v0 h 2m
(D)
(E)

28. DO NOW
A fire hose held near the ground shoots water at a speed of 6.6 m/s. At what angle(s) should the nozzle point in order that the water land 2.0 m away? Why are there 2 different angles?
q = 13.70, 76.30

29. EXAMPLE 300 600 a) t = 2.76 s b) Dymax = 9.30m c) range=64.5m
A player kicks a football from ground level with an initial velocity of 27 m/s at an angle of 300 above the horizontal.
Determine the ball’s hang time (total time in air)
Determine the ball’s maximum height
Determine the ball’s range
What is the ball’s impact velocity
e) The player then kicks the ball with the same speed but at a 60 degree angle. Which quantities change, which stay the same?
300
600
a) t = 2.76 s
b) Dymax = 9.30m
c) range=64.5m
d) v=27m/s -300
a) t = 4.78 s
b) Dymax = 27.9m
c) range=64.5m
d) v=27m/s -600 29

30. EXAMPLE- Projectile over Level ground
A player kicks a football from ground level with an initial velocity of v0 at an angle of q above the horizontal. The ball lands at the same height. Neglect air resistance.
Derive an expression for the ball’s hang time (in terms of v0, q, and g)
Derive an expression for the ball’s maximum height (in terms of v0, q, and g)
Derive an expression for the ball’s range (in terms of v0, q, and g) Use the trig identity 2sinqcosq = sin(2q)
Find an expression for the impact velocity
e) If the launch velocity of the football is 27 m/s, 300 above the horizontal, determine the ball’s hang time, max height, range and impact velocity.
f) Player then kicks the ball with the same speed but at a 60o angle. Which quantities change, which stay the same?
qo below the horizontal 30

31. EXAMPLE A movie stunt driver on a motorcycle speeds horizontally off a 50.0 m high cliff. How fast must the motorcycle leave the cliff top if it is to land on level ground below, 90.0 m from the base of the cliff where the cameras are? What is its final velocity, right before landing?
vo = 28.5 m/s
v0 = ? x y
50.0 m
90.0 m

32. EXAMPLE vf = 42.3 m/s, -47.70 from x-axis
What is its final velocity, right before landing?
vf = 42.3 m/s, from x-axis x y
50.0 m
vf = ?
90.0 m

33. DO NOW Dx = 37.9 m vf = 21 m/s, 45.40 below the x-axis
Frustrated with the book you are reading, you open the second story classroom window and violently hurl your book out the window with a velocity of 18 m/s at an angle of 35o above the horizontal. If the launch point is 6 m above the ground,
how far from the building will the book hit the parking lot?
What is the final velocity of the book right before it hits the ground?
Dx = 37.9 m
vf = 21 m/s, below the x-axis