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

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
$u t+\frac{1}{2} g t^{2}=\frac{1}{2} g n^{2}+g n t+\frac{1}{2} g t^{2}$
$t(u-g n)=\frac{1}{2} g n^{2}$
$t=\frac{1 / 2 g n^{2}}{u-g n} \ldots(i)$
$t$ - time at which second stone is thrown down
$\therefore $ Distance traveled $S=u t+1 / 2 g t^{2}$ by second stone
$S=t\left[u+\frac{1}{g} t\right]$
$S=\frac{1}{2} g n^{2}+g n t+1 / 2 g t^{2}$
$=\frac{g}{2}\left[n^{2}+2 n t+t^{2}\right]=\frac{g}{2}[n+t]^{2}$
$=\frac{g}{2}\left[n+\frac{1 / 2 g n^{2}}{u-g n}\right]^{2}$
$=\frac{g}{2}\left[\frac{m u-g n^{2}+\frac{g n^{2}}{2}}{4-g n}\right]^{2}$
$S=\frac{g}{2}\left[\frac{m u-\frac{g n^{2}}{2}}{(u-g n)}\right]^{2}=\frac{g}{2}\left[\frac{n\left(u-g \frac{u}{2}\right)}{(u-g n)}\right]^{2}$