A ball is thrown vertically downwards from a height of 20 m with an initial velocity v_0. It collides with the ground, loses 50 percent of its energy in collision and rebounds to the same height. The initial velocity v_0 is (take g = 10 ms^(-2))

A ball is thrown vertically downwards from a height of 20 m with an in

1.	A ball is thrown vertically downwards from a height of 20 m with an initial velocity v_0. It collides with the ground, loses 50 percent of its energy in collision and rebounds to the same height. The initial velocity v_0 is (take g = 10 ms^(-2))
Answer» `28 MS^(-1)` `10 ms^(-1)` `14 ms^(-1)` `20 ms^(-1)` Solution :The situation is shown in the figure. Let v 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., `= 1/2 mv^2 - 1/2 mv_0^2 = mgh` (where m is the mass of the ball) or `v^2 - v_0^2= 2gh "" .......(i)` Now , when the ball collides with the ground 50% of its energy is lost and it rebounds to the same height h. `:. (50)/(100) (1/2 mv^2) = mgh " or " I/4 v^2 = gh` or `v^2 = 4gh` Substituting this value of `v^2 ` in EQN. (i) we GET `4gh - v_0^2 = 2gh " or " v_0^2 = 4gh - 2gh = 2gh " or " v_0 = sqrt(2gh)` Here, `G = 10 ms^(-2) and h = 20 m` `:. v_0 = sqrt(2(10 ms^(-2))(20 m)) = 20 ms^(-1)`.