Question

A stone is allowed to fall from the top of a tower 100 m high. 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

Height of the tower = 100 m
$g=10m/s^2$
Velocity of projection of stone from the ground = 25 m/s
Velocity of projection of stone from the top of the tower = 0
Let the particles meet at time t and at height h from the ground.
Thus, height covered by the projectile projected from ground
$h=25t-\frac{1}{2}g{t}^2$ ...... (1)
Distance covered by the projectile thrown down in time t, $100-h=\frac{1}{2}g{t}^2$ ...... (2)
Adding equations (1) and (2),
$25t=100\Rightarrow t=4s$
Height at which the particles meet, $h=25t-5t^2=25\left(4\right)-5(4{)}^2=20m$