A ball of mass 0.2 ~kg is thrown vertically down from a height of 10 ~m . It collides with the floor and loses 50 % of its energy and then rises back to the same height. The value of its initial velocity is

Step-by-step solution

Given, mass of the ball, m=0.2 ~kg Height from the surface of floor, h=10 ~m Let initial velocity be u . Total energy as initial point TE = 1 2 m v^2+m g h After collision, remaining energy = 1 2 m u^2+m g h 2 With this remaining energy the ball bounces upto height h Therefore, 1 2 m u^2+m g h 2 =m g h aligned & 1 2 m u^2+m g h=2 m g h & aligned u^2 & = 2 g h & = 2 10 10 =14 ~m / s aligned aligned