Question

A stone is allowed to fall from the top of a tower 100m high and at the same time, another stone is projected vertically upwards from the ground with a velocity of 25m/s. Calculate when and where the two stones will meet.

Answer 1

The point of collision can be considered as x and the time of collision be t then we can find the x and t by using the two sets of data given. We can simplify things by realizing that there is no acceleration between the two stones.

Complete step by step answer
Let us say that two stones hit each other at distance x above the ground at time t, also let the point of collision is P, as shown in the diagram

For a stone A to reach at a point P, the downward acceleration is acting on that stone, so the displacement x and time t is related as $x = {u_A}t - \dfrac{1}{2}g{t^2}$………………. (1)
Here g is the acceleration due to gravity and uA is the given initial velocity
Similarly, for stone dropped from top, it has no downward velocity but only a downward acceleration. Also, we know that the tower is 100 m high. So, if stone A has travelled a distance x, then stone B has travelled the distance metre. So, the displacement of the stone B is given as $100 - x = \dfrac{1}{2}a{t^2}$ ……………………… (2)
Now, add equation (1) and (2), we get
$\Rightarrow 100 - x + x = {u_A}t + \dfrac{1}{2}a{t^2}$
$\Rightarrow 100 = {u_A}t$
Let us take the value of ${u_A}$ and ${u_B}$ to get the time taken.
$\Rightarrow 100 = 25t$
$\Rightarrow t = \dfrac{{100}}{{25}} = 4s$
Now, the position of the collision, x can be found by directly putting the value of t obtained in equation (1)
$\Rightarrow x = 25 \times 4 - \dfrac{1}{2} \times 9.8 \times {\left( 4 \right)^2}$
$\Rightarrow x = 100 - 78.4 = 21.6m$

Note
Here, both objects experience the same acceleration g in the same direction. So, we can say there is no relative acceleration between the two stones. So, from stone A perspective, stone B is approaching with zero acceleration.
Since, the relative velocity of stone A with respect to stone B is ${u_B} - {u_A} = 0 - 25 = - 25m{s^{ - 1}}$ towards stone B. with this in mind, we can easily calculate the time of collision. We have two bodies approaching each other at 125m/s with 100m separation. The time taken to collide would be $t = \dfrac{{dis\tan ce}}{{speed}} = \dfrac{{100}}{{25}} = 4s$