Physics - Kinematics Question with Solution | TestHub

The maximum height of a projectile thrown vertically upward is given by, $h=\frac{u^2}{2g}$.

$\Rightarrow u=\sqrt{2gh}$

Using equation of motion with constant acceleration,

$s=ut+\frac{1}{2}a{t}^2$

$\Rightarrow \frac{h}{3}=\sqrt{2gh}t-\frac{1}{2}g{t}^2$

$\Rightarrow \frac{g{t}^2}{2}-\sqrt{2ght}+\frac{h}{3}=0$

Solving the above quadratic equation we will get the times $t_1$ and $t_2$. $\Rightarrow \frac{t_2}{t_1}=\frac{\sqrt{2gh}+\sqrt{2gh-4\times \frac{g}{2}\times \frac{h}{3}}}{\sqrt{2gh}-\sqrt{2gh-4\times \frac{g}{2}\times \frac{h}{3}}}=\frac{\sqrt{2gh}+\sqrt{\frac{4gh}{3}}}{\sqrt{2gh}-\sqrt{\frac{4gh}{3}}}=\frac{\sqrt{3}+\sqrt{2}}{\sqrt{3}-\sqrt{2}}$

Thus, the ratio $\frac{t_1}{t_2}=\frac{\sqrt{3}-\sqrt{2}}{\sqrt{3}+\sqrt{2}}$