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
$h=-u{t}_1-\frac{1}{2}g{t}_1^2$ ..(i)
For downward motion $h=-u{t}_2+\frac{1}{2}g{t}_2^2$ ..(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 $\frac{1}{2}g{t}^2=\frac{1}{2}g{t}_1{t}_2$
$\Rightarrow$ $t^2=t_1{t}_2\Rightarrow t=\sqrt{t_1{t}_2}$