A ball is released from a height h . If t _ 1 and t _ 2 be the time...

As the body is dropped from a certain height $h$, its initial velocity will be zero.

For first half, ${h}_{1} = \dfrac{h}{2}$.

Using second equation of motion, $s = ut + \dfrac{1}{2}a{t}^{2}$

We have, $\dfrac{h}{2} = \dfrac{1}{2}g{t}_{1}^{2}$.

Time taken in travel first half of the height, ${t}_{1} = \sqrt{\dfrac{h}{g}} . . . \left(1\right)$.

For total height $h$, using again second equation of motion,

$h = \dfrac{1}{2}g{\left({t}_{1} + {t}_{2}\right)}$^{2} ⇒ $\left({t}_{1} + {t}_{2}\right) = \sqrt{\dfrac{2h}{g}} . . . \left(2\right)$

From equation \left(1\right)$and \left(2\right)$,

$⇒ \dfrac{1}{\sqrt{2}} = \dfrac{{t}_{1}}{{t}_{1} + {t}_{2}}$

$⇒ 1 + \dfrac{{t}_{2}}{{t}_{1}} = \sqrt{2}$

$⇒ \dfrac{{t}_{1}}{{t}_{2}} = \dfrac{1}{\sqrt{2} - 1}$

Thus, the relation is ${t}_{2} = \left(\sqrt{2} - 1\right){t}_{1}$.