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 = 27.0m/s and H = 25.0m.

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 = 27.0m/s and H = 25.0m. How far from the foot of the cliff does the ball land? How fast is it moving just before it lands?

Explanation / Answer

Time can be calculated as follows.

t = sqrt (2H/g) = sqrt (2*25/9.8) = 2.26 s

a)

We have that,

Distance = velocity * time = 27 * 2.26 = 61 m

b)

From the law of conservation of energy

       0.5mv^2 + mgh = 0.5mv'^2

         v^2 + 2gh = v'^2

          v'^2 = 27^2 + 2*9.8*25

            v' = 35 m/s

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