If a stone is dropped from height h

then \(h - \frac{1}{2} g t^{2}\) …(i)

If a stone is thrown upward with velocity u then

\(h = -vt_1 + \frac{1}{2} g t_1^2\) …(ii)

If a stone is thrown downward with velocity u then

\(\hbar - i \hbar \dot{c}_{2} + \frac{1}{2} g \dot{c}_{2} .\) …(iii)

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

\(- v_{1} + \frac{1}{2} g t^{2} - \frac{1}{2} g t^{2}\) …(iv)

\(\alpha_{x} + \frac{1}{2} g t^{2} - \frac{1}{2} g t^{2}\) …(v)

Dividing (iv) and (v) we get

. \(-\frac{u_1}{u_2} = \frac{\frac{1}{2}gt^2 - (r_1)}{\frac{1}{2}gt^2 - (r_2)}\)

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

By solving \(\tau = \sqrt{\tau_1 \tau_2}\)