Question

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 $\frac{h}{3}$ 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

Answer: (C)

$u=\sqrt{2gh}$
Now, $S=\frac{h}{3}a=-g$
$S=ut+\frac{1}{2}a{t}^2$
$\frac{h}{3}=\sqrt{2gh}t+\frac{1}{2}(-g)t^2$
$t^2\left(\frac{g}{2}\right)-\sqrt{2gh}t+\frac{h}{3}=0$
From quadratic equation
$t_1,t_2=\frac{\sqrt{2gh}\pm \sqrt{2gh-\frac{4gh}{2}\frac{h}{3}}}{g}$
$\frac{t_1}{t_2}=\frac{\sqrt{2gh}-\sqrt{\frac{4gh}{3}}}{\sqrt{2gh}+\sqrt{\frac{4gh}{3}}}$
$=\frac{\sqrt{3}-\sqrt{2}}{\sqrt{3}+\sqrt{2}}$