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 (Figure 1) . At the top of the hill, it is moving horizontally; then it goes over the vertical cliff. Take V = 29.0m/s and H = 21.0m .

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

B. How fast is it moving just before it lands?

C. Notice that when the ball lands, it has a larger translational speed than it had at the bottom of the hill. Does this mean that the ball somehow gained energy by going up the hill? Explain!

Explanation / Answer

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

E.bottom = 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= 25 m/s
PE = mgh
h = 0

E.top = KE + PE
KE = 7/10 m v2^2
PE = mgh
h = 28 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 = ?((25m/s) - 10/7*9.81m/s^2 * 28m)
v2 = 15.25 m/s

1. time to fall
h = 1/2 g t^2
t = ?(2h/g)
t = 2.39 s

Distance traveled:
L = t*v2
L = 36.45 m

2. v.vertical = ?(2gh)
v.vert = 23.4 m/s

total velocity = ? (v.hor^2 + v.vert^2)
= ? (15.25^2 + 23.4^2)
= 27.9 m/s

3. No, It is not rotating as fast. It has lost rotational energy.

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