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 = 23.0 m/s and H = 29.0 m

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

Express your answer in meters to three significant figures.

How fast is it moving just before it lands?

Express your answer in meters per second to three significant figures.

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

we use conservation of energy.

E.bottom = KE + PE
KE = 1/2 m v12 + 1/2 I 2
I = 2/5 m r2
= v/r
KE = 1/2m (v12 + 2/5 v12) = 7/10 m v12
v1= 23 m/s
PE = mgh
h = 0

E.top = KE + PE
KE = 7/10 m v22
PE = mgh
h = 29 m

E.bottom = E.top

7/10 m v12 + 0 = 7/10 m v22 + mg h
v22 = v12 -10/7gh
v2 = sqrt((23m/s)2 - 10/7*9.81m/s2 * 29m)
v2 = 11.072 m/s

1. time to fall
h = 1/2 g t^2
t = sqrt(2h/g)
t = 2.43 s

Distance traveled:
L = t*v2
L = 26.92 m

2. v.vertical = sqrt(2gh)
v.vert = 23.85 m/s

total velocity = sqrt(v.hor^2 + v.vert^2)
= sqrt(11.0722 + 23.852)
= 26.294 m/s

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

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