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Question: A stone is dropped from the top of a tall cliff and n seconds later another stone is thrown vertical...

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

g2[n(u−gn2)(u−gn)]2\frac{g}{2}\left\lbrack \frac{n\left( u - \frac{gn}{2} \right)}{(u - gn)} \right\rbrack^{2}

B

g2[n(u2−gn)(u−gn)]2\frac{g}{2}\left\lbrack \frac{n\left( \frac{u}{2} - gn \right)}{(u - gn)} \right\rbrack^{2}

C

g2[n(u2−gn)(u2−gn)]2\frac{g}{2}\left\lbrack \frac{n\left( \frac{u}{2} - gn \right)}{\left( \frac{u}{2} - gn \right)} \right\rbrack^{2}

D

g5[(u−gn)u2−gn]2\frac{g}{5}\left\lbrack \frac{(u - gn)}{\frac{u}{2} - gn} \right\rbrack^{2}

Answer

g2[n(u−gn2)(u−gn)]2\frac{g}{2}\left\lbrack \frac{n\left( u - \frac{gn}{2} \right)}{(u - gn)} \right\rbrack^{2}

Explanation

Solution

Let the two stones meet at time t.

For the first stone,

S1=12gt2S_{1} = \frac{1}{2}gt^{2} (∵u=0\because u = 0) …… (i)

For the seconds stone,

S2=u(t−n)+12g(t−n)2S_{2} = u(t - n) + \frac{1}{2}g(t - n)^{2} …….. (iii)

Displacement is same

∴S1=S2\therefore S_{1} = S_{2}

12gt2=u(t−n)+12g(t−n)2\frac{1}{2}gt^{2} = u(t - n) + \frac{1}{2}g(t - n)^{2} (Using (i) and (ii))

12gt2=ut−nu+12gt2+12gn2−gtnut−gtn=nu−12gn2\frac{1}{2}gt^{2} = ut - nu + \frac{1}{2}gt^{2} + \frac{1}{2}gn^{2} - gtnut - gtn = nu - \frac{1}{2}gn^{2}

t=nu−12gn2u−gn=n(u−g2n)u−gnt = \frac{nu - \frac{1}{2}gn^{2}}{u - gn} = \frac{n\left( u - \frac{g}{2}n \right)}{u - gn}

Substituting this value of t in (i), we get the required answer.