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.
Let the two stones meet after a time t.
(i) For the stone dropped from the tower:
Initial velocity, u= 0
Let the displacement of the stone in time t from the top of the tower be s.
Acceleration due to gravity, g = 9.8 m s- 2
From the equation of motion,
(ii) For the stone thrown upwards:
Initial velocity, u= 25 m s- 1
Let the displacement of the stone from the ground in time tbe s'.
Acceleration due to gravity, g = - 9.8 m s - 2
Equation of motion,
The combined displacement of both the stones at the meeting point is equal to
the height of the tower 100 m.
In 4 s, the falling stone has covered a distance given by equation (1) as
