Question

A stone dropped from a balloon which is at a height $ h, $ reaches the ground after t second. From the same balloon, if two stones are thrown, one upwards and the other downwards, with the same velocity $ u $ and they reach the ground after $ {{t}_{1}} $ and $ {{t}_{2}} $ second respectively, then

• A. $ t={{t}_{1}}-{{t}_{2}} $
• B. $ t=\frac{{{t}_{1}}+{{t}_{2}}}{2} $
• C. $ t=\sqrt{{{t}_{1}}{{t}_{2}}} $
• D. $ t=\sqrt{t_{1}^{2}-t_{2}^{2}} $

Answer 1

Answer: (C)

From equation of motion, we have
$ s=ut+\frac{1}{2}g{{t}^{2}} $ where, $ u $ is initial velocity, g the acceleration due to gravity and $ t $ the time.
For upward motion
$ h=-u{{t}_{1}}-\frac{1}{2}gt_{1}^{2} $ ..(i)
For downward motion $ h=-u{{t}_{2}}+\frac{1}{2}gt_{2}^{2} $ ..(ii)
Multiplying Eq. (i) by $ {{t}_{2}} $ and Eq. (ii) by $ {{t}_{1}} $ and subtracting Eq. (ii) by Eq. (i),
we get $ h({{t}_{2}}-{{t}_{1}})=\frac{1}{2}g{{t}_{1}}{{t}_{2}}({{t}_{2}}-{{t}_{1}}) $
$ h=\frac{1}{2}g{{t}_{1}}{{t}_{2}} $ ..(iii)
When stone is dropped from rest
$ u=0, $
reaches the ground in t second.
$ \therefore $ $ h=\frac{1}{2}g{{t}^{2}} $ ..(iv)
Equating Eqs. (iii) and (iv),
we get $ \frac{1}{2}g{{t}^{2}}=\frac{1}{2}g{{t}_{1}}{{t}_{2}} $
$ \Rightarrow $ $ t^2 = t_1 t_2 \Rightarrow t = \sqrt {t_1t_2}$