Question

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

• A. $\sqrt{3}v_{0}$
• B. $3\,v_0$
• C. $9\,v_0$
• D. $3/2\,v_0$

Answer 1

Answer: (A)

$\sqrt{3}v_{0}$

Answer 2

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 $v=0, u=v_{0}$ $0=v_{0}^{2}-2 g h$ $\Rightarrow v_{0}=\sqrt{2 g h}$ when $h'=3 h$ then $v_{0}'=\sqrt{2 g \times 3 h}=\sqrt{3} \sqrt{2 g h}=\sqrt{3} v_{0'}$ If ball has to be thrown to a greater height its initial velocity should be more than the original one.