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

Let after time t both stone meet and s be the distance travelled by the stone dropped from the top of tower at which the stones will meet. Distance travelled by the stone dropped = S
∴ Distance travelled by the stone projected upwards (100 - s)m
For the stone dropped from the tower.

(u = 0 because stone is dropped ie it starts from rest)
s = 5t² …………. (i)
For the stone projected upward
s = ut – \(\frac{1}{2}\) gt²
(Due to upward motion negative sign is taken)

The stones will meet after; 4s
s = 5t² = 5 × (4)² = 80 m
The stones will be at a distance of 80 m from the top of tower or 20 mt (100 m – 80 m) from the base of the tower 20m (100m – 80m).