A solid, uniform ball rolls without slipping up a hill, as shown in the figure.

Question

A solid, uniform ball rolls without slipping up a hill, as shown in the figure. At the top of the hill, it is moving horizontally; then it goes over the vertical cliff. Take V=29.0m/s and H=25.0m

Part A) How far from the foot of the cliff does the ball land?

Part B) How fast is it moving just before it lands?

A solid, uniform ball rolls without slipping up a hill, as shown in the figure. At the top of the hill, it is moving horizontally; then it goes over the vertical cliff. Take V=29.0m/s and H=25.0m Part A) How far from the foot of the cliff does the ball land? Part B) How fast is it moving just before it lands?

Explanation / Answer

You need to find the velocity at the top of the cliff. Use conservation of energy.

Ebottom = KE + PE
KE = 1/2 m v1^2 + 1/2 I ?^2
I = 2/5 m r^2
? = v/r
KE = 1/2m (v1^2 + 2/5 v1^2) = 7/10 m v1^2
v1 = 29 m/s
PE = mgh
h = 0

E.top = KE + PE
KE = 7/10 m v2^2
PE = mgh
h = 25 m

E.bottom = E.top

7/10 m v1^2 + 0 = 7/10 m v2^2 + mg h
v2^2 = v1^2 -10/7gh
v2 = ?((29m/s)2 - 10/7 * 9.81m/s^2 * 25m)
v2 = 22.15 m/s

1. time to fall
h = 1/2 g t^2
t = ?(2h/g)
t = ?(2*25/9.81)
t = 2.26 s

Distance traveled:
L = t*v2
L = 2.26 * 22.15
L = 50.01 m

2.
v.vertical = ?(2gh)
v.vertical  = ?(2*9.81*25)
v.vertical  = 22.15 m/s

total velocity = ? (v.hor^2 + v.vert^2)
= ? (22.15^2 + 22.15^2)
total velocity = 31.32 m/s

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