A ball falls under gravity from a height 10 m with an initial velocity `v_0`. It hits the ground, loses 50% of its energy in collision and it rises to the same height, what is the value of `v_0`? 

Options

A. 14 m/s
B. 7 m/s
C. 28 m/s
D. 9.8 m/s

Answer

A

Explanation

Let v be the velocity of the ball when it hits the ground.
then `v^2 = v_0^2 + 2gh = v_0^2 + 2 xx 9.8 xx 10`
or `v^2 = v_0^2 + 196`
Let `v.` be the velocity after impect and it rise same height 10 m.
`:. V.^(2) - 0 = 2 xx 9.8 xx 10 = 196 or v. = 14 m//s`
Ratio of kinetic energy before impact and after impact
`(1/2 mv^2)/(1/2 mv.^2) = (v^2)/(v.^2) = (v_0^2 + 196)/(196) = 2 " or " v_0 = 14 m//s` .