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. `sqrt3 V_0`
B. `3v_0`
C. `9V_0`
D. `3//2V_0`

Answer

A

Explanation

`Since `H_(max) = (u^2)/(2g) rArr H_(max) prop u^2`
`u prop sqrt(H_(max))`
`u_1/u_2 = sqrt((H_(max))/(3H_(max))) rArr v_2 = sqrt3u_1 i.e., sqrt3v_0` .