When a ball is thrown up vertically with velocity v0, it reaches a maximum height of h. If one wishes to triple the maximum height then the ball should be thrown with velocity:

Question

When a ball is thrown up vertically with velocity v0, it reaches a maximum height of h. If one wishes to triple the maximum height then the ball should be thrown with velocity: (A) √3 v0 (B) 3 v0 (C) 9 v0 (D) 3 / 2 v0

Tardigrade Admin · Accepted Answer

At maximum height velocity is zero.
From equation of motion we have

v^{2} = u^{2} - 2 g h

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}

\Rightarrow v_{0} = 2 g h

when

h^{'} = 3 h

then

v_{0}^{'} = 2 g \times 3 h = 3 2 g h = 3 v_{0}

Q. 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:

$3 v_{0}$

$3 v_{0}$

$9 v_{0}$

$3/2 v_{0}$

Q. When a ball is thrown up vertically with velocity v0​, it reaches a maximum height of h. If one wishes to triple the maximum height then the ball should be thrown with velocity:

3​v0​

3v0​

9v0​

3/2v0​

Q. 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:

$3 v_{0}$

$3 v_{0}$

$9 v_{0}$

$3/2 v_{0}$