Question

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 ground in $t_2$ seconds. At what time will it reach the ground if it is dropped from A?

• A. $\sqrt{t_1/t_2}$
• B. $\sqrt{t_2/t_1}$
• C. $\sqrt{t_1t_2}$
• D. $t_1 t_2$

Answer 1

Answer: (C)

For downward motion,
$h = ut_1 + \frac{1}{2} gt_1^2 \,\,...(i)$
For upward motion,
$h = -ut_2 + \frac{1}{2} gt_2^2\,\,...(ii)$
From $(i)$ & $(ii)$
$ut_1 + \frac{1}{2} gt_1^2 = - ut_2 + \frac{1}{2} gt^2_2$
$\Rightarrow u(t_1 + t_2) = \frac{1}{2} g(t_2^2 - t_1^2)$
$\therefore h = \frac{g}{2} (t_1 - t_1)t_1 + \frac{g}{2} gt_1^2 = \frac{gt_2t_1}{2}$
$\therefore $ Required time $ t = \sqrt{\frac{2h}{g}} $
$= \sqrt{t_1t_2}$.