A ball is thrown vertically downwards from a height of 20m with an intial velocity `v_(0)`. It collides with the ground, loses 50% of its energy in collision and rebounds to the same height. The intial velocity `v_(0)` is (Take, g =10 `ms^(-2)`)

Options

A. `28 ms^(-1)`
B. `10 ms^(-1)`
C. `14 ms^(-1)`
D. `20 ms^(-1)`

Answer

`(4)`

Explanation

Let v be the velocity of the ball with which it collides with ground. Then according to the law of conservation of energy, Gain 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)
`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.
`therefore 50/100(1/2 mv^(2))` = mgh
`1/4 v^(2) = gh or v^(2) = 4gh`
Substituting this value of `v^(2)` in eqn.(i) , we get
` 4gh - v_(0)^(2) = 2gh`