A ball is thrown vertically downwards from a height of 20 m with an initial velocity ν0. It collides with the ground, loses 50 percent of its energy in collision and rebounds to the same height. The initial velocity ν0 is (Take g = 10 ms-2 )

The situation is shown in the figureLet ν be the velocity of the ball with which it collides with ground. Then according to the law of conservation of energy,Gain in kinetic energy = loss in potential energy  i.e. 12mv2-12mv02=mgh (where m is the mass of the ball)  or   v2-v02=2gh              . . . .(i)Now, when the ball collides with the ground, 50% of its energy is lost and it rebounds to the same height h.∴  5010012mv2=mgh14v2=gh   or   v2=4gh Substituting this value of v2 in eqn. (i), we get 4gh-v02=2gh or   v02=4gh-2gh=2gh or v0=2gh Here, g=10 ms-2and h=20 m∴  v0=210 ms-2(20 m)=20 ms-1