Question

When a ball is thrown up vertically with velocity $v_o$ 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 $\sqrt{3}{v}_0$

Maximum height means final velocity is zero.
Thus $H=\frac{0-u^2}{-2g}=\frac{u^2}{2g}$
To triple maximum height, i.e. 3H, we have, $3H=\frac{{u^{\prime }}^2}{2g}$ or ${u^{\prime }}^2=3u^2$ or, $u^{\prime }=\sqrt{3}u$