I have solved this question below from my textbook. My answer differs from the one given in the book. I have tried my best to see where I am wrong. I need your help to find the truth of this question.

Q. Two stones are thrown up simultaneously from the edge of a $200\,\mathrm{m}$ high cliff with initial speeds of $15\,\mathrm{m.s^{-1}}$ and $30\,\mathrm{m.s^{-1}}$. The graph representing the time variation of the relative position of the second stone with respect to the first stone would be linear for 8 seconds. Is it true or false?

I think it is false and the graph would be linear for $1.5$ seconds only when the first stone reaches its peak but the second stone continues upward journey.

The answer in the text book is $8\,\mathrm{s}$.


closed as off-topic by John Rennie, Dilaton, Abhimanyu Pallavi Sudhir, Kyle Kanos, jinawee Feb 11 '14 at 15:23

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Relative position versus time is linear when relative velocity is constant i.e. when relative acceleration eqauls zero. In this case, both stones are acted upon by gravity only and hence their acceleration always equals $g$ directed downwards. Therefore, relative position varies linearly for the entire duration of motion which stops when the stone thrown with $15$ m/s reaches the ground. To calculate that

$-200 = 15t - 0.5gt^2$ which on solving gives $t = 8$ sec

So, the textbook is absolutely right!!!

  • $\begingroup$ thats the explanation I needed very much! thanks a lot.. $\endgroup$ – KawaiKx Feb 12 '14 at 1:27

Hint: The speed of the object number $i$ ($i=1$ or $2$) will be $$v_i(t)=v_i(0)-gt.$$


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