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)

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
$v=0, u=v_0$
$\Rightarrow v_0=\sqrt{2 g h}$
when $h^{\prime}=3 h$ then
$v_0^{\prime}=\sqrt{2 g \times 3 h}=\sqrt{3} \sqrt{2 g h}=\sqrt{3} v_0$