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

\frac{g}{2} {⎡ ⎢ ⎢ ⎢ ⎢ ⎣ \frac{n (u - \frac{g n}{2})}{u - g n} ⎤ ⎥ ⎥ ⎥ ⎥ ⎦}^{2}

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\frac{g}{2} {⎡ ⎢ ⎢ ⎢ ⎢ ⎣ \frac{n (\frac{u}{2} - g n)}{u - g n} ⎤ ⎥ ⎥ ⎥ ⎥ ⎦}^{2}

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\frac{g}{2} {⎡ ⎢ ⎢ ⎢ ⎢ ⎣ \frac{n (\frac{u}{2} - g n)}{\frac{u}{2} - g n} ⎤ ⎥ ⎥ ⎥ ⎥ ⎦}^{2}

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\frac{g}{2} {⎡ ⎢ ⎢ ⎣ \frac{(u - g n)}{\frac{u}{2} - g n} ⎤ ⎥ ⎥ ⎦}^{2}

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Solution

The correct option is A $\frac{g}{2} {⎡ ⎢ ⎢ ⎢ ⎢ ⎣ \frac{n (u - \frac{g n}{2})}{u - g n} ⎤ ⎥ ⎥ ⎥ ⎥ ⎦}^{2}$
$S = \frac{1}{2} g t^{2} . . . . . . . . (1)$
And that of the other stone is
$S = u (t - n) + \frac{1}{2} [g (t - n)^{2}] . . . . . . . . (2)$

Since both the stones meet at the distance so equation (1)and equation(2) will be equal
$\frac{1}{2} g t^{2} = u (t - n) + \frac{1}{2} [g (t - n)^{2}]$

$g t^{2} = 2 u t - 2 u n + g t^{2} + g n^{2} - 2 g n t$

$t (2 g n - 2 u) = g n^{2} - 2 u n$

$t = \frac{n (\frac{g n}{2} - u)}{(g n - u)}$

now putting value of t in equation(1)

$s = \frac{g}{2} {⎡ ⎢ ⎢ ⎣ \frac{[n (\frac{g n}{2} - u)]}{{g n - u}^{2}} ⎤ ⎥ ⎥ ⎦}^{2}$

$S = \frac{g}{2} {⎡ ⎢ ⎢ ⎣ \frac{n (\frac{g n}{2} - u)}{(g n - u)^{2}} ⎤ ⎥ ⎥ ⎦}^{2}$

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