QJEEM22Ph2e8a: JEE Main 2022 Physics - Class 11 JEEM22Ph2e8a Solution: A ball is thrown up vertically with a certain velocity so that, …

A ball is thrown vertically upward with a certain velocity, reaching a maximum height $h$ . We need to find the ratio of the times it is at height $\frac{h}{3}$ while ascending and descending, respectively.

The initial velocity of the ball $v$ can be given by:

$v = \sqrt{2gh}$

When the ball is at height $\frac{h}{3}$ , the equation of motion is:

$\frac{h}{3} = \sqrt{2gh} \cdot t - \frac{1}{2}gt^2$

Rearranging the equation, we get a quadratic equation in terms of $t$ :

$\frac{g}{2} t^2 - \sqrt{2gh} \cdot t + \frac{h}{3} = 0$

The ratio of the times taken to ascend and descend to the height $\frac{h}{3}$ can be found using the quadratic formula solution for $t$ , considering the corresponding velocities:

$\frac{t_1}{t_2} = \frac{\sqrt{2gh} + \sqrt{2gh - \frac{2gh}{3}} }{\sqrt{2gh} - \sqrt{2gh - \frac{2gh}{3}}}$

Simplifying the terms inside the fraction:

$= \frac{\sqrt{2} + \frac{2}{\sqrt{3}}}{\sqrt{2} - \frac{2}{\sqrt{3}} } = \frac{\sqrt{3} + \sqrt{2}}{\sqrt{3} - \sqrt{2}}$