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 ms^-1. Calculate when and where the two stones will meet.

Given,

(a) In the case, when the stone falls from the top of the tower,

Initial velocity u = 0

Distance travelled = x

Time taken = t

According to the second equation of motion,

S = ut + $\dfrac{1}{2}$gt²

⇒ S = 0 x t + $\dfrac{1}{2}$gt²

⇒ S = $\dfrac{1}{2}$gt²   [Eq 1]

(b) When another stone is projected vertically upwards,

Initial velocity u = 25 ms^-1

Distance travelled = (100 - x)

Time taken = t

Using S = ut + $\dfrac{1}{2}$gt²

we get,

S' = 25t - $\dfrac{1}{2}$gt²   [Eq 2]

From equations (1) and (2)

S + S' = $\dfrac{1}{2}$gt² + 25t - $\dfrac{1}{2}$gt²

100 = $\dfrac{1}{2}$gt² + 25t - $\dfrac{1}{2}$gt²

100 = 25t

t = $\dfrac{100}{25}$ = 4 sec.

Hence, after 4 secs, the two stones will meet.

From (1)

S = $\dfrac{1}{2}$ x 10 x 4² = 5 x 16 = 80m.

Hence, after 4 sec, 2 stones meet a distance of 80 m from the top.