A ball is thrown up vertically with a certain velocity so that, it...

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

$⇒ u = \sqrt{2gh}$

Using equation of motion with constant acceleration,

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

$⇒ \dfrac{h}{3} = \sqrt{2gh}t - \dfrac{1}{2}g{t}^{2}$

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

Solving the above quadratic equation we will get the times ${t}_{1}$ and ${t}_{2}$. $⇒ \dfrac{{t}_{2}}{{t}_{1}} = \dfrac{\sqrt{2gh} + \sqrt{2gh - 4 \times \dfrac{g}{2} \times \dfrac{h}{3}}}{\sqrt{2gh} - \sqrt{2gh - 4 \times \dfrac{g}{2} \times \dfrac{h}{3}}} = \dfrac{\sqrt{2gh} + \sqrt{\dfrac{4gh}{3}}}{\sqrt{2gh} - \sqrt{\dfrac{4gh}{3}}} = \dfrac{\sqrt{3} + \sqrt{2}}{\sqrt{3} - \sqrt{2}}$

Thus, the ratio $\dfrac{{t}_{1}}{{t}_{2}} = \dfrac{\sqrt{3} - \sqrt{2}}{\sqrt{3} + \sqrt{2}}$