A body is projected vertically downwards from A, the top of the tower, reaches the ground in $t_1$ seconds. If it is projected upwards with same velocity, it reaches the top in $t_2$ seconds. At what time will it reach the ground if it is dropped from A?

For downward motion,

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

For upward motion,

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

when dropped from A

$\mathrm{h}=\frac{1}{2}{\mathrm{gt}}^2 ........\left(\mathrm{iii}\right)$

From (i) & (ii)

${\mathrm{ut}}_1+\frac{1}{2}{\mathrm{gt}}_1^2=-{\mathrm{ut}}_2+\frac{1}{2}{\mathrm{gt}}_2^2$

$\Rightarrow u(t_1+t_2)=\frac{1}{2}g(t_2^2-t_1^2)\Rightarrow u =\frac{1}{2} g(t$₂ - t₁)

Substituting in equation (i),

$\therefore h=\frac{g}{2}(t_2-t_1)t_1+\frac{g}{2}g{t}_1^2=\frac{g{t}_2{t}_1}{2}$

for dropped body,

$\therefore Requiredtimet=\sqrt{\frac{2h}{g}}=\sqrt{t_1{t}_2}$