# A body is dropped from a certain point and it travels a height ‘h’ after some time. What would be the time required to travel double the height?

Let the body travel the height $h$ in time $t$. Since the body is dropped, initial velocity is zero $(u=0).$

$h=ut+\frac{1}{2}gt^2$ $h=0+\frac{1}{2}gt^2$ $\therefore t=\sqrt{\frac{2h}{g}}$

If the new height is doubled the original, $h’=2h.$

Let $t’$ be the new required time. $t’=\sqrt{\frac{2h’}{g}}=\sqrt{\frac{2(2h)}{g}}=\sqrt{2}\sqrt{\frac{2h}{g}}=\sqrt{2}\;t$

Thus, the new time will be $\sqrt{2}$ times the original time.

[Equations used above are deduced from Motion Under Gravity]