A stone is thrown upwards from the top of a tower with some initial speed and it reaches the ground in 16 seconds. Now it is thrown with the same initial speed downward and it reaches the ground in 9 sec. In how much time will it reach the ground if the stone is allowed to fall freely under gravity from the same place ?

Suppose the body is projected vertically upwards from A with a speed ${\mathrm{u}}_{\mathrm{o}}$

Using equation s = ut +$\left(\frac{1}{2}\right){\mathrm{at}}^2$

For first case, $-\mathrm{h} = {\mathrm{u}}_{\mathrm{o}}{\mathrm{t}}_1-\left(\frac{1}{2}\right){{\mathrm{gt}}_1}^2 ----\left(\mathrm{i}\right)$

For second case, $\mathrm{A} - \mathrm{h} = -{\mathrm{u}}_{\mathrm{o}}{\mathrm{t}}_2 -\left(\frac{1}{2}\right){{\mathrm{gt}}_2}^2----\left(\mathrm{ii}\right)$

(i) -(ii) $0 = {\mathrm{u}}_0({\mathrm{t}}_2+{\mathrm{t}}_1)+\left(\frac{1}{2}\right)\mathrm{g}({{\mathrm{t}}_{2 }}^2-{{\mathrm{t}}_1}^2)$

${\mathrm{u}}_0 = \left(\frac{1}{2}\right)\mathrm{g}({\mathrm{t}}_1-{\mathrm{t}}_2)-----\left(\mathrm{iii}\right)$

Put the value of ${\mathrm{u}}_0 \mathrm{in} \left(\mathrm{ii}\right),$

$-\mathrm{h} = -\left(\frac{1}{2}\right)\mathrm{g}({\mathrm{t}}_1-{\mathrm{t}}_2){\mathrm{t}}_2-\left(\frac{1}{2}\right){{\mathrm{gt}}^2}_2$

$\Rightarrow \mathrm{h} = \frac{1}{2}{\mathrm{gt}}_1{\mathrm{t}}_2$------(iv)

For third case, u = 0, t = ?

$-\mathrm{h} = 0\times \mathrm{t}-\left(\frac{1}{2}\right){\mathrm{gt}}^2 \mathrm{or} \mathrm{h} = \left(\frac{1}{2}\right){\mathrm{gt}}^2----\left(\mathrm{v}\right)$

Combining equation (iv) and equation (v), we get

$\frac{1}{2}{\mathrm{gt}}^2 = \frac{1}{2}{\mathrm{gt}}_1{\mathrm{t}}_2 \mathrm{or} \mathrm{t} = \sqrt{{\mathrm{t}}_1{\mathrm{t}}_2}$

Putting the value we get t = 12 sec