Projectile Motion Examples

Example 3-6: Driving off a cliff!!

y is positive upward, y0 = 0 at top. Also vy0 = 0

vx = vx0 = ? vy = -gtx = vx0t, y = - (½)gt2

Time to Bottom: t = √2y/(-g) = 3.19 svx0 = (x/t) = 28.2 m/s

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 to land on level ground below, 90.0 m from the base of the cliff where the cameras are?

Example 3-7: Kicked football

A football is kicked at an angle θ0 = 37.0° with a velocity of 20.0 m/s, as shown.

Calculate:

a. Max height. b. Time when hits ground. c. Total distance traveled in the x direction. d. Velocity at top. e. Acceleration at top.

θ0 = 37º, v0 = 20 m/s

vx0= v0cos(θ0) = 16 m/s, vy0= v0sin(θ0) = 12 m/s

Conceptual Example 3-8

• Demonstration!!

v0x

Conceptual Ex. 3-9: Wrong Strategy

• “Shooting the Monkey”!!

• Video Clip!!

Example 3-10

• Range (R) of projectile Maximum horizontal distance before returning to ground. Derive a formula for R.

• 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)

Example 3-11, A punt!

• v0 = 20 m/s, θ0 = 37º

• vx0= v0cos(θ0) = 16 m/s, vy0= v0sin(θ0) = 12 m/s

Proof that 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

Of the form y = Ax – Bx2

A parabola in the x-y plane!!

Ex. 3-12 a): Rescue Helicopter Drops SuppliesA 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?

A rescue helicopter wants to get a package of supplies to isolated mountain climbers on a rocky ridge 200 m below. The helicopter is traveling horizontally with a speed of 70 m/s (250 km/h), b) 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?

Ex. 3-12 b), c): Rescue Helicopter Throws Supplies

That’s Quite an Arm!Problem: A stone is thrown

from the top of a building at an angle θ0 = 26° to the horizontal and with an 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?

Example: The Long Jump

ProblemA long-jumper leaves the ground at angle θ0 = 20° above the horizontal

at a speed of v0 = 8.0 m/s.

a. How far does he jump in the horizontal direction?B. What is his maximum height?

Stranded ExplorersProblem: 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

Problem 46

Chapter 3, Problem 46 SolutionChoose the origin at ground level, under where the projectile is launched, & up to be the positive y direction. For the projectile:

a. The time to reach the ground is found from Eq. 2-12b, with final height = 0. Choose positive time since the projectile was launched at time t = 0.

b. The horizontal range is found from the horizontal motion at constant velocity.

0 65.0 m s ,v 0 35.0 , ,ya g 0 115m,y 0 0 0sin .yv v

2 21 10 0 0 02 2

2 2 120 0 0 0

12

0

0

0 sin

sin sin 49.964s , 2.3655s 9.96s

2

y yy y v t a t y v t gt

v v gt

g

y

0 0cos 65.0 m s cos35.0 9.964s 531mxx v t v t

0 0cos 65.0 m s cos35.0 53.2 m s .xv v

c. At the instant just before the particle reaches the ground, the horizontal component of its velocity is the constant

The vertical component of velocity is found from Eq. 2-12a:

d. The magnitude of the velocity is found from the x and y components calculated in part c. above.

2

0 0 0sin 65.0 m s sin 35.0 9.80 m s 9.964s

60.4 m s

y yv v at v gt

0 0cos 65.0 m s cos35.0 53.2 m s .xv v

2 22 2 53.2 m s 60.4 m s 80.5m sx yv v v

e. The direction of the velocity is

so the object is moving

f. The maximum height above the cliff top reached by the projectile will occur when the y-velocity is 0, and is found from Eq. 2-12c.

1 1 60.4tan tan 48.6

53.2y

x

v

v

48.6 below the horizon .

2 2 2 2

0 0 0 0 max

2 22 2

0 0max 2

2 0 sin 2

65.0 m s sin 35.0sin70.9 m

2 2 9.80 m s

y y yv v a y y v gy

vy

g