A ball is thrown up with a certain velocity so that it reaches a height ' h '.

$\mathbf{u}=\sqrt{2 \mathrm{gh}}$

Now,

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

$\mathrm{S}=\mathrm{ut}+\frac{1}{2} \mathrm{at}^{2}$

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

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

From quadratic equation

$\mathrm{t}_{1}, \mathrm{t}_{2}=\frac{\sqrt{2 \mathrm{gh}} \pm \sqrt{2 \mathrm{gh}-\frac{4 \mathrm{~g} \mathrm{}}{2} \frac{\mathrm{h}}{3}}}{\mathrm{~g}}$

$\begin{aligned} \frac{t_{1}}{t_{2}} &=\frac{\sqrt{2 g h}-\sqrt{\frac{4 g h}{3}}}{\sqrt{2 g h}+\sqrt{\frac{4 g h}{3}}} \\ &=\frac{\sqrt{3}-\sqrt{2}}{\sqrt{3}+\sqrt{2}} \end{aligned}$