1. Projectile Motion Examples

2. Demonstration next time!!
Conceptual Example
v0x 
Demonstration next time!!

3. Conceptual Example: Wrong Strategy
“Shooting the Monkey”!!
Video Clip next time!!

4. Example
Range (R) of the projectile  Maximum horizontal distance before returning to ground. Derive a formula for R.

5. Range R  the x where y = 0!
Use vx = vx0 , x = vx0 t , vy = vy0 - gt
y = vy0 t – (½)g t2, (vy) 2 = (vy0)2 - 2gy

6.  t = 0 (of course!) and t = (2vy0)/g
Range R  the x where y = 0!
Use vx = vx0 , x = vx0 t , vy = vy0 - gt
y = vy0 t – (½)g t2, (vy) 2 = (vy0)2 - 2gy
First, find the time t when y = 0
0 = vy0 t - (½)g t2
 t = 0 (of course!) and t = (2vy0)/g

7.  t = 0 (of course!) and t = (2vy0)/g
Range R  the x where y = 0!
Use vx = vx0 , x = vx0 t , vy = vy0 - gt
y = vy0 t – (½)g t2, (vy) 2 = (vy0)2 - 2gy
First, find the time t when y = 0
0 = vy0 t - (½)g t2
 t = 0 (of course!) and t = (2vy0)/g
Put this t in the x formula: x = vx0 (2vy0)/g  R
R = 2(vx0vy0)/g

8. vx0= v0cos(θ0), vy0= v0sin(θ0) R = (v0)2 [2 sin(θ0)cos(θ0)]/g
Range R  the x where y = 0!
Use vx = vx0 , x = vx0 t , vy = vy0 - gt
y = vy0 t – (½)g t2, (vy) 2 = (vy0)2 - 2gy
First, find the time t when y = 0
0 = vy0 t - (½)g t2
 t = 0 (of course!) and t = (2vy0)/g
Put this t in the x formula: x = vx0 (2vy0)/g  R
R = 2(vx0vy0)/g
vx0= v0cos(θ0), vy0= v0sin(θ0)
R = (v0)2 [2 sin(θ0)cos(θ0)]/g
R = (v0)2 sin(2θ0)/g
(by a trig identity)

9. Example: A punt! v0 = 20 m/s, θ0 = 37º vx0= v0cos(θ0) = 16 m/s
vy0= v0sin(θ0) = 12 m/s

10. Proof that the projectile path is a parabola
x = vx0 t , y = vy0 t – (½)g t2
Note: The same time t enters both equations!
 Eliminate t to get y as a function of x.
Solve x equation for t: t = x/vx0
Get: y = vy0 (x/vx0) – (½)g (x/vx0)2
Or: y = (vy0 /vx0)x - [(½)g/(vx0)2]x2
This means that it is of the form
y = Ax – Bx2
A parabola in the x-y plane!!

11. Example: Rescue Helicopter Drops Supplies
A rescue helicopter wants to drop a package of supplies to isolated
mountain climbers on a rocky ridge 200 m below. If the helicopter is
traveling horizontally with a speed of 70 m/s (250 km/h), a) How far in
advance of the recipients (horizontal distance) must the package be
dropped? b) Instead, someone in the helicopter throws the package
a horizontal distance of 400 m in advance of the mountain
climbers. What vertical velocity should the package be given (up or
down) so that it arrives precisely at the climbers’ position? c) With
what speed does the package land in the latter case?
Figure 3-29.
Answer: a. We can find the time the package will take to fall, and then the horizontal distance it will travel: 450 m
b. We know the horizontal distance and the horizontal speed (the speed of the helicopter), so we can find the time the package will take to fall. Substituting gives the initial y velocity: -7.0 m/s (downward)
c. 94 m/s

12. Problem

13. b) What is the speed of the stone just before it strikes the ground?
Example: That’s Quite an Arm!
Problem: A stone is thrown from the top of a building at angle θ0 = 26° to the horizontal and with initial speed v0 = 17.9 m/s, as in the figure. The height of the building is 45.0 m.
a) How long is the stone
"in flight"?
b) What is the speed of the stone just before it strikes the ground?

14. Example: Stranded Explorers
Problem: An Alaskan rescue plane drops a package of emergency rations to a stranded party of explorers, as shown in the picture. If the plane is traveling horizontally at v0 = 42.0 m/s at a height h = 106 m above the ground, where does the package strike the ground relative to the point at which it is released?
v0 = 42 m/s h