Question

A ball is thrown up with certain velocity so that it reaches a height $h$. Find the ratio of the times in which it is at $\frac{h}{3}$.

• A. $\frac{\sqrt{2}-1}{\sqrt{2}+1}$
• B. $\frac{\sqrt{3}-\sqrt{2}}{\sqrt{3}+\sqrt{2}}$
• C. $\frac{\sqrt{3}-1}{\sqrt{3}+1}$
• D. $\frac{1}{3}$

Answer 1

The correct option is B $\frac{\sqrt{3}-\sqrt{2}}{\sqrt{3}+\sqrt{2}}$

At maximum height, velocity is zero so, $0=u^2-2ghoru=\sqrt{2gh}$
As $s=ut+\frac{1}{2}a{t}^2$ so $,\frac{h}{3}=\sqrt{2gh}t-\frac{1}{2}g{t}^2$
$g{t}^2-2\sqrt{2gh}t+\frac{2h}{3}=0ort=\frac{2\sqrt{2gh}\pm \sqrt{8gh-\frac{8gh}{3}}}{2g}$
$\frac{t_1}{t_2}=\frac{2\sqrt{2gh}-\sqrt{8gh-\frac{8gh}{3}}}{2\sqrt{2gh}+\sqrt{8gh-\frac{8gh}{3}}}=\frac{\sqrt{3}-\sqrt{3-1}}{\sqrt{3}+\sqrt{3-1}}=\frac{\sqrt{3}-\sqrt{2}}{\sqrt{3}+\sqrt{2}}$