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 = 30.0m/s and H = 20.0m .

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

How fast is it moving just before it lands

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

7/10*vx^2 + 9.8*20 = 7/10*30^2

vx = 24.9 m/s

y= 1/2 * g*t^2

20 = 1/2*9.8*t^2

t = 2.02 s

a)

x = 24.9*2.02 = 50.3 m

b)

vy = 9.8*2.02 = 19.8 m/s

v = sqrt(vx^2 + vy^2) = sqrt(24.9 ^2 + 19.8^2)

v = 31.81 m/s

Notice that this is larger than the initial speed! The ball is spinning more slowly, so more energy is available for linear motion.

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