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,
$S_{1}=\frac{1}{2} gt^{2}\quad\left(\because u=0\right)\quad\ldots\left(i\right)$
For the second stone,
$S_{2}=u\left(t-n\right)+\frac{1}{2}g\left(t-n\right)^{2}\quad\ldots\left(ii\right)$
Displacement is same $\therefore S_{1}=S_{2}$
$\frac{1}{2} gt^{2}=u\left(t-n\right)+\frac{1}{2}g\left(t-n\right)^{2}\quad$ (Using $(i)$ and $(ii)$)
$\frac{1}{2}gt^{2}=ut-nu+\frac{1}{2}gt^{2}+\frac{1}{2}gn^{2}-gtn$
$ut-gtn=nu-\frac{1}{2} gn^{2}$
$t=\frac{nu-\frac{1}{2}gn^{2}}{u-gn}=\frac{n\left(u-\frac{g}{2}n\right)}{u-gn}$ or $S_{1}=\frac{g}{2}\left[\frac{n\left(u-g \frac{n}{2}\right)}{\left(u-gn\right)}\right]^{2}\quad$ [from $(i)$]