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 stone will meet.

Answer

4s 80 m from the top.

Explanation

Let t be the point at which two stones meet and let h be their height from the ground Height of the tower is H= 100 m (Given).
It is clear from the question that we need to calculate time when the two stones met. After calculating time, we will also be able to calculate the distance.
Now, first consider the stone which falls from the top of the tower.
Initial velocity (u)= 0
So, distance covered by this stone at time t can be calculated using equation of motion
`x-x_(0)=u_(0)t+1/2gt^(2)`
Since initial velocity u=0 so we get
`100-x=1/2gt^(2)...(1)`
The distance covered by the same stone that is thrown in upward direction from ground is
`x=u_(0)t-1/2gt^(2)`
In this case initial velocity is 25 m/s. So,
`x=25t-1/2gt^(2)....(2)`
Adding equations (1) and (2) we get,
100=25t
Or, t=4s
Putting value in equation (2),
`x=25xx4-1/2xx9.8xx(4)^(2)=100-78.4=21.6 cm`