Question

A body projected vertically upwards with a certain speed from the top of a tower reaches the ground in . If it is projected vertically downwards from the same point with the same speed, it reaches the ground in . 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:

$t = \sqrt{t_1 t_2}$