When a ball is thrown up vertically with velocity `v_0`, it reaches a maximum height of h. If one wishes to triple the maximum height then the ball should be thrown with velocity

Options

A. `sqrt(3) v_0`
B. `3 v_0`
C. `9 v_0`
D. `3//2 v_0`

Answer

A

Explanation

At maximum height velocity is zero.
From equation of motion we have
`v^2 = u^2 - 2gh`
where `v` is final velocity, `u` is initial velocity. Since ball reaches maximum height, velocity at the highest point is zero. Therefore, we have
`v = 0, u = v_0`
`rArr v_0 = sqrt(2 gh)`
when `h' = 3 h` then
`v'_0 = sqrt(2g xx 3h) = sqrt(3) sqrt(2 gh) = sqrt(3) v_0`.