A ball is projected vertically down with an initial velocity from a height of 20 m onto a horizontal floor. During the impact it loses 50% of its energy and rebounds to the same height. The initial velocity of its projection is

Let ball is projected vertically downward with velocity v from height hTotal energy at point A = 12mv2+mghDuring collision loss of energy is 50% and the ball rises up to same height. It means it possess only potential energy at same level.50%12mv2+mgh=mgh1212mv2+mgh=mghv = 2gh = 2×10×20v = 20 m/s