Motion Under Gravity question from JEE Main 2022 (29 Jul, Shift 1)

Answer: (b)

1. Maximum height $h$ gives $u^2 = 2gh$.
2. At height $\frac{h}{3}$: $\frac{h}{3} = ut - \frac{1}{2}gt^2 \Rightarrow t = \frac{u \pm \sqrt{u^2 - \frac{2gh}{3}}}{g}$
3. $u = \sqrt{2gh}$ and $\sqrt{u^2 - \frac{2gh}{3}} = \sqrt{\frac{4gh}{3}}$
4. $\frac{t_1}{t_2} = \frac{\sqrt{2} - \sqrt{4/3}}{\sqrt{2} + \sqrt{4/3}} = \frac{\sqrt{6} - 2}{\sqrt{6} + 2} = \frac{\sqrt{3} - \sqrt{2}}{\sqrt{3} + \sqrt{2}}$
5. So the answer is (b) $\frac{\sqrt{3}-\sqrt{2}}{\sqrt{3}+\sqrt{2}}$.