Question

A ball is thrown up with a certain velocity so that it reaches a height ${}^{\prime }{h}^{\prime }$. Find the ratio of the two different times of the ball reaching $\frac{h}{3}$ in both the directions.

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

Answer 1

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

Using the conservation of energy theorem,

Projected speed, $u=\sqrt{2gh}$

Now, when ball is moving upward

$S=\frac{h}{3}:a=-g$

Using, $S=ut+\frac{1}{2}a{t}^2$

$\frac{h}{3}=\sqrt{2gh}t+\frac{1}{2}(-g)t^2$

$t^2\left(\frac{g}{2}\right)-\sqrt{2gh}t+\frac{h}{3}=0$

From quadratic equation,

$t_1,t_2=\frac{\sqrt{2gh}\pm \sqrt{2gh-\frac{4g}{2}\frac{h}{3}}}{g}$

So, the ratio of two different times of the ball reaching $h/3$ is

$\frac{t_1}{t_2}=\frac{\sqrt{2gh}-\sqrt{\frac{4gh}{3}}}{\sqrt{2gh}+\sqrt{\frac{4gh}{3}}}=\frac{\sqrt{6}-\sqrt{4}}{\sqrt{6}+\sqrt{4}}$

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

Hence, the correct option is $\left(\text{C}\right)$.