## University Physics with Modern Physics (14th Edition)

From a height of $3h$, it would take a time of $\sqrt{3} ~T$ to reach the ground.
Let $h$ be the height where it was dropped originally. $h = \frac{1}{2}gT^2$ $T = \sqrt{\frac{2h}{g}}$ Now suppose the object is dropped from a height of $3h$. $3h = \frac{1}{2}gt_2^2$ $t_2 = \sqrt{\frac{6h}{g}}$ $t_2 = \sqrt{3}\times \sqrt{\frac{2h}{g}} = \sqrt{3} ~T$ From a height of $3h$, it would take a time of $\sqrt{3} ~T$ to reach the ground.