Question

related to work power energy..plz only post answer if u really know it bcoz due irrelevant posting of answers the relevant answers cant be posted n this really creates a problem......thus requested hope users understand

Answer 1

applying conservation of energy..

$mgh \: + \: \frac{1}{2} m {v}^{2} \: = mgr$
v - initial speed of the ball
solving this equation,
$v \: = \sqrt{2g(r - h)}$
this is the answer.

Answer 2

Work energy theorem
1/2 mv^2+ mgh - mgR=0
v= root( 2g(R-h))