Question

From the top of a tall mountain, a stone is dropped. One second later, another stone is projected downwards with velocity $20{\text{ms}}^{-1}$. Find the distance from top where both stones meet.

• A. $\frac{45}{4}\text{m}$
• B. $40\text{m}$
• C. $10\text{m}$
• D. $\frac{50}{8}\text{m}$

Answer 1

The correct option is A $\frac{45}{4}\text{m}$

The distance covered by first stone in $1\text{s}$,

$S_1=\frac{1}{2}g{t}^2=\frac{1}{2}\left(10\right)(1{)}^2=5\text{m}$

Velocity of the first stone after $1\text{s}$,

$v_1=u+at(\text{here},a=g)$

$v_1=0+10\left(1\right)=10{\text{ms}}^{-1}$

At this instant, the velocity of second stone,

$v_2=20\text{m/s}$ (At the instant of release)

Now using concept of relative motion,

$v_{rel}=v_2-v_1=10{\text{ms}}^{-1}$

$S_{rel}=5\text{m};a_{rel}=0$

Time taken to second stone to cross first stone after throw is given by

$t=\frac{S_{rel}}{v_{rel}}=\frac{5}{10}=\frac{1}{2}\text{s}$

Time from $t=0$ to time of crossing for first stone will be

$T=1+\frac{1}{2}=\frac{3}{2}\text{s}$

So the total distance covered by first stone is

$S_{total}=\frac{1}{2}g{T}^2=\frac{1}{2}\times 10\times {\left(\frac{3}{2}\right)}^2$

$\therefore S_{total}=\frac{45}{4}\text{m}$

Hence, option $\left(\text{A}\right)$ is the correct answer.