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:
$u = u, v = 0$
Let $t_{1}$ be the time of ascent. Using the equation of motion $v = u + at$:
$0 = u - gt_{1}$
$t_{1} = u/g$
For downward motion (descent):
The body starts from rest at the maximum height,so $u = 0$. Let $t_{2}$ be the time of descent. Using $s = ut + (1/2)at^{2}$ where $s = h$:
$h = 0 + (1/2)gt_{2}^{2} \implies t_{2} = \sqrt{2h/g}$
Since $h = u^{2}/(2g)$,substituting this gives $t_{2} = \sqrt{2(u^{2}/2g)/g} = u/g$.
Thus,$t_{1} = t_{2}$,which proves that the time of ascent is equal to the time of descent.