Question

show that time of ascent=time of descent

Answer 1

a = -g --> v = -gt + v_0. Tada! We just derived the first basic kinematics equation with integration. Now, we see that the time of ascent is equal to (v_0-v)/g and the time of ascent is equal to (v-v_0)/(-g). Same thing. Boom!

Answer 2

the absolut height as a function of time at vertical throw is:
\[s(t) = -\frac{ 1 }{ 2 }*g*(t - t _{maximum})^{2} + s _{maximum}\]

now, there exist two time-differences between the time at any height and the time of maximum...
\[t - t _{maximum} = \pm \sqrt{-2*\frac{ s(t) - s _{maximum} }{ g }}\]
...which's magnitudes are the same, as you can see!