A stone dropped from a building of height h and it reaches after t seconds on earth. From the same building if two stones are thrown (one upwards and other downwards) with the same velocity u and they reach the earth surface after t₁ and t₂  seconds respectively, then

If a stone is dropped from height h, then

$h=\frac{1}{2}g{t}^2 ...... \left(i\right)$

If a stone is thrown upward with velocity u then

$h=-u{t}_1+\frac{1}{2}g{t}_1^2 ............. \left(ii\right)$

If a stone is thrown downward with velocity u then

$h=u{t}_2+\frac{1}{2}g{t}_2^2 ........... \left(iii\right)$

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

$h=u{t}_2+\frac{1}{2}g{t}_2^2 .......... \left(iv\right)$

$u{t}_2+\frac{1}{2}g{t}_2^2=\frac{1}{2}g{t}^2 ........ \left(v\right)$

Dividing (iv) and (v) we get

$\therefore \frac{-u{t}_1}{u{t}_2}=\frac{\frac{1}{2}g\left(t^2-t_1^2\right)}{\frac{1}{2}g\left(t^2-t_2^2\right)}$

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

$\text{ By solving, }t=\sqrt{t_1{t}_2}$