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When the ball is at h the PE1 of the ball is mgh

When the ball is pushed with a velocity v it  reaches the bottom with KE1=1/2mv^2

Then by inertia the ball travels upto R when its PE2= mgR and the velocity =0 KE2=0

From law of conservation of energy

PE1+KE1=PE2+KE2 or mgh+1/2mv^2=mgR+0 solving it we get V= √2g(R-h)

Answer

When the ball is at h the PE1 of the ball is mgh

When the ball is pushed with a velocity v it  reaches the bottom with KE1=1/2mv^2

Then by inertia the ball travels upto R when its PE2= mgR and the velocity =0 KE2=0

From law of conservation of energy

PE1+KE1=PE2+KE2 or mgh+1/2mv^2=mgR+0 solving it we get V= √2g(R-h)

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