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

$u = \sqrt{2gh}$
Now,

$S = \dfrac{h}{3}, a = - g$

$S = ut + \dfrac{1}{2}a{t}^{2}$
$\dfrac{h}{3} = \sqrt{2gh}t + \dfrac{1}{2}\left(- g\right){t}^{2}$
${t}^{2}\left(\dfrac{g}{2}\right) - \sqrt{2gh}t + \dfrac{h}{3} = 0$

From quadratic equation

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

$\dfrac{{t}_{1}}{{t}_{2}} = \dfrac{\sqrt{2gh} - \sqrt{\dfrac{4gh}{3}}}{\sqrt{2gh} + \sqrt{\dfrac{4gh}{3}}}$

$= \dfrac{\sqrt{3} - \sqrt{2}}{\sqrt{3} + \sqrt{2}}$