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

Options

A. `20ms^(-1)`
B. `15ms^(-1)`
C. `10ms^(-1)`
D. `5ms^(-1)`

Answer

A

Explanation

Let ball is projected vertically downward with velocity v from height h.
Total energy at point `A = 1/2mv^(2) + mgh`
During collision, loss of energy is 50% and the ball rise up to same height. It means only potential energy at same level.
`50%(1/2mv^(2) + mgh) = mgh`
`1/2 (1/2mv^(2) + mgh) = mgh`
`v = sqrt(2gh)` = `sqrt(2 xx 10 xx 20)`
`v=20ms^(-1)`