Question

A stone is dropped from the top of 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. $\frac{g}{2}{\left[\frac{n\left(u-\frac{gn}{2}\right)}{(u-gn)}\right]}^2$
• B. $\frac{g}{2}{\left[\frac{n\left(\frac{u}{2}-gn\right)}{(u-gn)}\right]}^2$
• C. $\frac{g}{2}\left[\frac{n\left(\frac{u}{2}-gn\right)}{\left(\frac{u}{2}-gn\right)}\right]$
• D. $\frac{g}{5}\left[\frac{(u-gn)}{\left(\frac{u}{2}-gn\right)}\right]$

Answer 1

Answer: (A)

Distance traveled by stone in n time
$u=O$ $t=n$
$\therefore S=\frac{1}{2}g{n}^2$
Equating the distance traveled by both stones when one stone is overtaking the other
$ut+\frac{1}{2}g{t}^2=\frac{1}{2}g{n}^2+gnt+\frac{1}{2}g{t}^2$
$t(u-gn)=\frac{1}{2}g{n}^2$
$t=\frac{1/2g{n}^2}{u-gn}\dots (i)$
$t$ - time at which second stone is thrown down
$\therefore$ Distance traveled $S=ut+1/2g{t}^2$ by second stone
$S=t\left[u+\frac{1}{g}t\right]$
$S=\frac{1}{2}g{n}^2+gnt+1/2g{t}^2$
$=\frac{g}{2}\left[n^2+2nt+t^2\right]=\frac{g}{2}[n+t{]}^2$
$=\frac{g}{2}{\left[n+\frac{1/2g{n}^2}{u-gn}\right]}^2$
$=\frac{g}{2}{\left[\frac{mu-g{n}^2+\frac{g{n}^2}{2}}{4-gn}\right]}^2$
$S=\frac{g}{2}{\left[\frac{mu-\frac{g{n}^2}{2}}{(u-gn)}\right]}^2=\frac{g}{2}{\left[\frac{n\left(u-g\frac{u}{2}\right)}{(u-gn)}\right]}^2$