A stone dropped from a building of height h reaches the ground after t seconds. From the same building if two stones are thrown (one upwards and the other downwards) with the same velocity$u$ and they reach the ground after $t_1$ and $t_2$ seconds respectively, then the time interval $t$ is

For the stone dropped with zero initial velocity. We have

$h=\frac{1}{2}g{t}^2$ …………………(i)

For the stone thrown upwards with velocity u, we have

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

For the stone thrown downwards with velocity u, we have

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

Notice that the displacement of the stone in the three case is the same, equal to h. Using (i) to (ii) and (iii) we have

$\frac{1}{2}g{t}^2=-u{t}_1+\frac{1}{2}g{t}_1{ }^2\Rightarrow u{t}_1=\frac{1}{2}g{t}_1{ }^2-\frac{1}{2}g{t}_{ }^2$

$\frac{1}{2}g{t}^2=u{t}_2+\frac{1}{2}g{t}_2{ }^2\Rightarrow u{t}_2=\frac{1}{2}g{t}_{ }{ }^2-\frac{1}{2}g{t}_2{ }^2$

Dividing the two equations ,we have

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

Which gives $t=\sqrt{t_1{t}_2}$.