Question

A stone is dropped from the top of a tall cliff and $n$ seconds later another stone is thrown vertically downwards with a velocity $u$. Then the second stone overtakes the first, below the top of the cliff at a distance given by

• A. $\frac{g}{2}{\left[\frac{n\left(u-\frac{gn}{2}\right)}{\left(u-gn\right)}\right]}^2$
• B. $\frac{g}{2}{\left[\frac{n\left(\frac{u}{2}-gn\right)}{\left(u-gn\right)}\right]}^2$
• C. $\frac{g}{2}{\left[\frac{n\left(\frac{u}{2}-gn\right)}{\left(\frac{u}{2}-gn\right)}\right]}^2$
• D. $\frac{g}{5}{\left[\frac{\left(u-gn\right)}{\left(\frac{u}{2}-gn\right)}\right]}^2$

Answer 1

Answer: (A)

Let the two stones meet at time $t$.
For the first stone,
$S_1=\frac{1}{2}g{t}^2\left(\because u=0\right)\dots \left(i\right)$
For the second stone,
$S_2=u\left(t-n\right)+\frac{1}{2}g{\left(t-n\right)}^2\dots \left(ii\right)$
Displacement is same $\therefore S_1=S_2$
$\frac{1}{2}g{t}^2=u\left(t-n\right)+\frac{1}{2}g{\left(t-n\right)}^2$ (Using $(i)$ and $(ii)$)
$\frac{1}{2}g{t}^2=ut-nu+\frac{1}{2}g{t}^2+\frac{1}{2}g{n}^2-gtn$
$ut-gtn=nu-\frac{1}{2}g{n}^2$
$t=\frac{nu-\frac{1}{2}g{n}^2}{u-gn}=\frac{n\left(u-\frac{g}{2}n\right)}{u-gn}$ or $S_1=\frac{g}{2}{\left[\frac{n\left(u-g\frac{n}{2}\right)}{\left(u-gn\right)}\right]}^2$ [from $(i)$]