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. $3v_0$
• C. $9v_0$
• D. $\frac{3}{2}{v}_0$

Answer 1

The correct option is A $\sqrt{3}{v}_0$

$\frac{1}{2}m{v}_0^2=mgh......\left(i\right)$
$\frac{1}{2}m{v}^2=mg\times 3h.....\left(ii\right)$
Dividing $\frac{v^2}{v_0^2}=3;v^2=3v_0^2$
$\Rightarrow v=\sqrt{3}{v}_0$