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))`

Options

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

Answer

D

Explanation

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)`.