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

The correct option is B $t=\sqrt{t_1{t}_2}$

Step 1, Given data

Height of the building = h

Time taken to reach the ground = t

Two other balls are thrown in upward and downward directions simultaneously.

Step 2, Finding the relation
Taking downward as positive
For first case of dropping the stone, $h=\frac{1}{2}g{t}^2$
For second case of downward throwing $h=u{t}_1+\frac{1}{2}g{t}_1^2=\frac{1}{2}g{t}^2$

Or we can write
$⟹u{t}_1=\frac{1}{2}g(t^2-t_1^2)$......(i)
For third case of throwing upwards $h=-u{t}_2+\frac{1}{2}g{t}_2^2=\frac{1}{2}g{t}^2$

Or we can write
$⟹u{t}_2=\frac{1}{2}g(t_2^2-t^2)$....(ii)
On solving (i) and (ii) we get,
$\frac{t_1}{t_2}=\frac{t^2-t_1^2}{t_2^2-t^2}$

Or,
$⟹t=\sqrt{t_1{t}_2}$

Hence the relation is as given above.