Question

A stone $A$ is dropped from rest from height $h$ above ground. A second stone $B$ is simultaneously thrown vertically up with velocity $v$. Find $v$ so that $B$ meets midway between their initial positions.

• A. $\sqrt{gh}$
• B. $2\sqrt{gh}$
• C. $3\sqrt{gh}$
• D. $4\sqrt{gh}$

Answer 1

The correct option is A $\sqrt{gh}$

Let time of travel of each stone $=t$
Distance travelled by each stone $=\frac{h}{2}$
Taking upward direction as positive
Using second equation of motion,
For stone $A$, $-\frac{h}{2}=-\frac{1}{2}{gt}^2$
$\Rightarrow t=\sqrt{\frac{h}{g}}$
For stone $B$, $\frac{h}{2}=vt-\frac{1}{2}g{t}^2$
Putting $t=\sqrt{\frac{h}{g}}$, we get
$v=\sqrt{gh}$

Alternate sol:
For stone $B$ w.r.t $A$,
$S_{rel}=h$, $u_{rel}=v-0=v\text{m/s},a_{rel}=0$
Using second equation of motion,
$S_{rel}=u_{rel}t+\frac{1}{2}a{t}^2$ $\Rightarrow h=vt$ $\Rightarrow t=\frac{h}{v}\dots \dots \left(1\right)$
For the body dropped from height $h$,
$\frac{h}{2}=\frac{1}{2}g{t}^2\Rightarrow t=\sqrt{\frac{h}{g}}\dots \dots \left(2\right)$
From (1) and (2) $\sqrt{\frac{h}{g}}=\frac{h}{v}\Rightarrow v=\sqrt{hg}$