then $h = \frac{1}{2}g\,{t^2}$…(i)

If a stone is thrown upward with velocity $u$ then

$h = - u\;{t_1} + \frac{1}{2}g\;t_1^2$…(ii)

If a stone is thrown downward with velocity $u$ then

$h = u{t_2} + \frac{1}{2}gt_2^2$…(iii)

From (i) (ii) and (iii) we get

$ - u{t_1} + \frac{1}{2}g\,t_1^2 = \frac{1}{2}g\,{t^2}$…(iv)

$u{t_2} + \frac{1}{2}g\,t_2^2 = \frac{1}{2}g\,{t^2}$…(v)

Dividing (iv) and (v) we get

$\therefore \frac{{ - u{t_1}}}{{u{t_2}}} = \frac{{\frac{1}{2}g({t^2} - t_1^2)}}{{\frac{1}{2}g({t^2} - t_2^2)}}$

or $ - \frac{{{t_1}}}{{{t_2}}} = \frac{{{t^2} - t_1^2}}{{{t^2} - t_2^2}}$

By solving $t = \sqrt {{t_1}{t_2}} $