Question

Question 17
A stone is allowed to fall from the top of a tower 100 m high and at the same time another stone is projected vertically upwards from the ground with a velocity of 25 m/s. Calculate when and where the two stones will meet.

Answer 1

Let t be the time at which two stones meet and let $h^{\prime }$ be their height from the ground.
Given that height of the tower, H = 100m
First consider the stone which falls from the top of the tower. Distance covered by this stone in time t can be calculated using the second equation of motion.
$h=ut+\frac{1}{2}g{t}^2$
Since initial velocity u = 0 so we get
$h=\frac{1}{2}g{t}^2\dots \dots \left(1\right)$
The distance covered by the stone that is thrown vertically upwards is
$h^{\prime }=ut-\frac{1}{2}g{t}^2$
In this case, initial velocity is 25 m/s. So,
$h^{\prime }=25t-\frac{1}{2}g{t}^2\dots \dots \left(2\right)$
Adding equations (1) and (2) we get,
h + h' = 25t
But h + h' = H = 100 m
So, 100 = 25 t
or, t = 4 s
Putting value in equation (2),
$h^{\prime }=25\times 4-\frac{1}{2}\times 9.8\times (4{)}^2=100-78.4=21.6m$
They will meet at a height of 21.6 m from ground.