Question

A stone $A$ is dropped from rest from height $h$ above the ground. A second stone $B$ is simultaneously thrown vertically up from a point on the ground with velocity $v$. Find $v$ so that $B$ meets $A$ 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 $=\frac{h}{2}$
Using second equation of motion,
Taking upward as positive for both stones,
For stone $A,-\frac{h}{2}=-\frac{1}{2}g{t}^2\Rightarrow t=\sqrt{\frac{h}{g}}$
For stone $B,\frac{h}{2}=vt-\frac{1}{2}g{t}^2$
Putting $t=\sqrt{\frac{h}{g}}$, we get
$\frac{h}{2}=v\sqrt{\frac{h}{g}}-\frac{1}{2}g{\left(\sqrt{\frac{h}{g}}\right)}^2$
$v=\sqrt{gh}$


Alternate Solution:
For stone $B$ w.r.t $A$,
$S_{rel}=S_B-S_A=\frac{h}{2}-(-\frac{h}{2})=h$
$u_{rel}=v-0=v\text{m/s}$ and $a_{rel}=0$
Using second equation of motion,
$S_{rel}=u_{rel}t+\frac{1}{2}a{t}^2$
$\Rightarrow h=vt$
$\Rightarrow t=\frac{h}{v}$ ... (1)
For stone $A$:
$\frac{h}{2}=\frac{1}{2}g{t}^2\Rightarrow t=\sqrt{\frac{h}{g}}$ ... (2)
From (1) & (2),
$\sqrt{\frac{h}{g}}=\frac{h}{v}\Rightarrow v=\sqrt{gh}$