Question

A ball is thrown up vertically with a certain velocity so that, it reaches a maximum height h. Find the ratio of the times in which it is at height $\frac{h}{3}$ while going up and coming down respectively.

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

Answer 1

Answer: (B)

$\text{ Max. Height }=h=\frac{u^2}{2g}$
$\Rightarrow u=\sqrt{2gh}$
$S=ut+\frac{1}{2}a{t}^2$
$\frac{h}{3}=\sqrt{2ght}+\frac{1}{2}(-g)t^2$
$\frac{g{t}^2}{2}-\sqrt{2ght}+\frac{h}{3}=0$
(Roots are $t_1\&t_2$ )
$\frac{t_2}{t_1}=\frac{\sqrt{2gh}+\sqrt{2gh-4\times \frac{g}{2}\times \frac{h}{3}}}{\sqrt{2gh}-\sqrt{2gh-4\times \frac{g}{2}\times \frac{h}{3}}}$
$=\frac{\sqrt{2gh}+\sqrt{\frac{4gh}{3}}}{\sqrt{2gh}-\sqrt{\frac{4gh}{3}}}=\frac{\sqrt{3}+\sqrt{2}}{\sqrt{3}-\sqrt{2}}$