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

Max. Hight $=\mathrm{h}=\frac{{\mathrm{u}}^2}{2\mathrm{g}}$
$\begin{array}{l}\Rightarrow \mathrm{u}=\sqrt{2\mathrm{gh}}\\ \mathrm{S}=\mathrm{ut}+\frac{1}{2}{\mathrm{at}}^2\\ \Rightarrow \frac{\mathrm{h}}{3}=\sqrt{2\mathrm{gh}}\mathrm{t}+\frac{1}{2}(-\mathrm{g}){\mathrm{t}}^2\\ \Rightarrow \frac{{\mathrm{gt}}^2}{2}-\sqrt{2\mathrm{gh}}\mathrm{t}+\frac{\mathrm{h}}{3}=0\end{array}$

Time taken while going up is t₁

Time taken while going up is t₂
(Roots are t₁ & t₂)
$\frac{{\mathrm{t}}_1}{{\mathrm{t}}_2}=\frac{\sqrt{2\mathrm{gh}}-\sqrt{2\mathrm{gh}-4\times \frac{\mathrm{g}}{2}\times \frac{\mathrm{h}}{3}}}{\sqrt{2\mathrm{gh}}+\sqrt{2\mathrm{gh}-4\times \frac{\mathrm{g}}{2}\times \frac{\mathrm{h}}{3}}}=\frac{\sqrt{2\mathrm{gh}}-\sqrt{\frac{4\mathrm{gh}}{3}}}{\sqrt{2\mathrm{gh}}+\sqrt{\frac{4\mathrm{gh}}{3}}}=\frac{\sqrt{3}-\sqrt{2}}{\sqrt{3}+\sqrt{2}}$