Prove that if a body is thrown vertically upward,the time of ascent is equal to the time of descent.

(N/A) For upward motion:
$v = u - gt$
At the highest point,final velocity $v = 0$.
So,$0 = u - gt_1$,which gives $t_1 = \frac{u}{g}$ $....(1)$
For downward motion:
The body starts from rest at the highest point,so initial velocity $u' = 0$.
Using $s = ut + \frac{1}{2}at^2$,where $s = h$ (maximum height) and $a = g$:
$h = 0 + \frac{1}{2}gt_2^2 \implies t_2 = \sqrt{\frac{2h}{g}}$.
Since $h = \frac{u^2}{2g}$,substituting this gives $t_2 = \sqrt{\frac{2(u^2/2g)}{g}} = \sqrt{\frac{u^2}{g^2}} = \frac{u}{g}$ $....(2)$
Comparing $(1)$ and $(2)$,$t_1 = t_2$. Thus,the time of ascent is equal to the time of descent.