A body is released from height h above the ground. Exactly at the midway if g vanishes suddenly, its total time of fall is                        

Given that

A body id dropped from height $=\text{H}$

$\text{u}=0$

(i) Final velocity$v=?$  

After descending through$\text{h}=\frac{\text{H}}{\text{2}}$  

We have ${\text{v}}^{\text{2}}=\text{2gh}$   

${\text{v}}^{\text{2}}=2\text{g}\frac{\text{H}}{\text{2}}$

$\therefore \text{v}=\sqrt{\text{gH}}$

(ii) Time taken to descend through$\frac{\text{H}}{\text{2}}$  

We have$\text{h =}\frac{1}{2}{\text{gt}}^{\text{2}}$  

$\frac{\text{H}}{\text{2}}=\frac{1}{2}{\text{gt}}^{\text{2}}$

$\therefore \text{t}=\sqrt{\frac{\text{H}}{\text{g}}}$  

Now acceleration due to gravity g becomes zero

The body will continue to descend with uniform velocity$\text{v}=\sqrt{\text{gH}}$  

Then time taken to cover$\text{h}=\frac{\text{H}}{\text{2}}$  

We know${\text{t}}^{\text{1}}=\frac{\text{s}}{\text{v}}=\frac{\frac{\text{H}}{\text{2}}}{\sqrt{\text{gH}}}$

$\therefore {\text{t}}^{\text{1}}=\frac{1}{2}\sqrt{\frac{\text{H}}{\text{g}}}$

Therefore the total time taken to reach the ground

$=\text{t}+{\text{t}}^{\text{1}}=\sqrt{\frac{\text{H}}{\text{g}}}+\frac{1}{2}\sqrt{\frac{\text{H}}{\text{g}}}$ $=\frac{3}{2}\sqrt{\frac{\text{H}}{\text{g}}}$