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)

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