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= 25.0 m/s and h = 25.0 m. 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= 25.0 m/s and h = 25.0 m. AA m AAAAaasdp[kferf[pea.) How far from the foot of the cliff does the ball land? How fast is it moving just before it lands?

Explanation / Answer

Moment of Inertia of solid sphere

I = (2/5)MR^2

Total Kinetic Energy KE = Rotational KE1 + Translational KE2

KE = KE1 + KE2

KE = (1/2)MV^2 + (1/2)I W^2 = (1/2)MV^2 + (1/2) (2/5)MR^2 (V/R)^2

KE = (7/10)MV^2

At top of the cliff , let the velocity be V1

Conservation Of Energy will hold good.

MgH + (7/10)MV1^2 = (7/10)MV^2

V1 = 16.58 m/s

The time taken by ball to fall the cliff is given by H = (1/2)gt^2

t = 2.26 Second

The ball will fall V1*t distance from foot of the cliff = 37.47 m

The vertical component of velocity due to gravity will be V2 = gt

Speed = sqrt(V1^2 + V2^2) = 27.68 m/s

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