Question

A body projected vertically upwards with a certain speed from the top of a tower reaches the ground in t1. If it is projected vertically downwards from the same point with the same speed, it reaches the ground in t2. Time required to reach the ground, if it is dropped from the top of the tower, is :

• A. $\sqrt{t_1 t_2}$
• B. $\sqrt{t_1 - t_2}$
• C. $\sqrt{\frac{t_1}{t_2}}$
• D. $\sqrt{t_1 + t_2}$

Answer 1

Answer: (A)

$\sqrt{t_1 t_2}$

Answer 2

Given:
$t_1 = \frac{u + \sqrt{u^2 + 2gh}}{g}$
$t_2 = \frac{-u + \sqrt{u^2 + 2gh}}{g}$
For the time $t$ required if the body is dropped (i.e., initial velocity $u = 0$):
$t = \sqrt{\frac{2gh}{g^2}} = \frac{\sqrt{2gh}}{g}$
Now, using the equations for $t_1$ and $t_2$:
$t_1 t_2 = \frac{(u^2 + 2gh) - u^2}{g^2} = \frac{2gh}{g^2} = t^2$
Thus:
$t = \sqrt{t_1 t_2}$