Question

A stone is dropped from the top of a tall cliff and n seconds later another stone is thrown vertically downwards with velocity $u$. Then the second stone overtakes the first, below the top of the cliff at a distance given by:
[Assume $u$ sufficiently enough to overtake the first stone]
(A). $\dfrac{g}{2}{\left[ {n\dfrac{{\left( {\dfrac{{gn}}{2} - u} \right)}}{{gn - u}}} \right]^2}$
(B). $\dfrac{g}{2}{\left[ {n\dfrac{{\left( {gn - \dfrac{u}{2}} \right)}}{{gn - u}}} \right]^2}$
(C). $\left( g \right){\left[ {\dfrac{{\left( {gn - u} \right)}}{{gn - \left( {\dfrac{u}{2}} \right)}}} \right]^2}$
(D). $\left( {\dfrac{g}{5}} \right){\left[ {\dfrac{{\left( {gn - u} \right)}}{{gn - \left( {\dfrac{u}{2}} \right)}}} \right]^2}$

Answer 1

Hint: Both the stone is going through the same path and one is under free fall and another is thrown with some velocity so obviously they will meet and the stone with the velocity will overtake the other stone. At that time the distance travelled by both the stones will be the same.

Complete step-by-step solution -
Let us assume a situation when both the stones meet each other and one stone which is thrown overtakes the other one. Let this phenomenon occur after $t$ seconds and the height at which they will overtake is $h$.
Now by using the following formula for the calculation of the distance travelled by both stones at the instance of overtaking,
$s = ut + \dfrac{1}{2}a{t^2}$
Now one by one, we will be using this formula for both the stones.
So for the first stone,
$ - h = - \dfrac{1}{2}g{(n + t)^2}$
$ \Rightarrow h = \dfrac{1}{2}g{(n + t)^2}$-------equation (1)
Similarly for the second stone we will write the expression for the displacement,
$ - h = - ut - \dfrac{1}{2}g{t^2}$
$ \Rightarrow h = ut + \dfrac{1}{2}g{t^2}$---------equation (2)
Now equating equations (1) and equation (2), we will get the following,
$\dfrac{1}{2}g{(n + t)^2} = ut + \dfrac{1}{2}g{t^2}$
Now after simplifying the above equation which we have got, we will get the time, which is as follows.
$t = \dfrac{{g{n^2}}}{{2\left( {u - gn} \right)}}$
Now we will put this value in the equation (1)
$h = \dfrac{g}{2}{[n + \dfrac{{g{n^2}}}{{2\left( {u - gn} \right)}}]^2}$
Finally we have,
$h = \dfrac{g}{2}{\left[ {n\dfrac{{\left( {\dfrac{{gn}}{2} - u} \right)}}{{gn - u}}} \right]^2}$
Hence the option (A) is the correct answer.

Note: If an object falls from a height and it is falling only under gravity and no external forces are applied then this is the condition of the free fall and in this condition the initial velocity of the object will be taken zero while if a object is thrown vertically downward with some force then its initial velocity will not be considered zero.