Q 11-02-102JEE MainJEE Main 2021 (27 Jul, Shift 1)Medium
A ball is thrown up with a certain velocity so that it reaches a height $h$. Find the ratio of the two different times of the ball reaching $\dfrac{h}{3}$ in both the directions.
Answer: (C) $\dfrac{\sqrt3-\sqrt2}{\sqrt3+\sqrt2}$
For maximum height $h$: $u^2 = 2gh$.
At height $\dfrac h3$: $\dfrac h3 = ut - \dfrac12 gt^2 \Rightarrow gt^2 - 2ut + \dfrac{2h}{3} = 0$
$$t = \frac{u \pm \sqrt{u^2 - \frac{2gh}{3}}}{g} = \frac{u}{g}\left(1 \pm \sqrt{\tfrac23}\right)$$
$$\frac{t_1}{t_2} = \frac{1 - \sqrt{2/3}}{1 + \sqrt{2/3}} = \frac{\sqrt3 - \sqrt2}{\sqrt3 + \sqrt2}$$
Solution by Sreeraj P, M.Sc Physics