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.

When the stone dropped from the tower

Initial velocity, u = 0 m/s

Let the displacement of the stone in time t from the top of the tower be s.

Acceleration due to gravity, g = 9.8 ms⁻²

From the equation of motion,

s = ut +{1\over2}at²

s = 0 × t +{1\over2} × 9.8 × t²

⇒ s = 4.9t² — (1)

When the stone thrown upwards

Initial velocity, u = 25 ms⁻¹

Let the displacement of the stone from the ground in time t be s′.

Acceleration due to gravity, g = −9.8 ms⁻²

Equation of motion,

s = ut +{1\over2}at²

s ′ = 25 × t − {1\over2} × 9.8 × t²

⇒ s ′ = 25t − 4.9t² — (2)

The combined displacement of both the stones at the meeting point is equal to the height of the tower 100 m.

s ′ + s = 100

⇒ 25t − 4.9t² + 4.9t² = 100

⇒ t = {100\over25} s = 4s

In 4 s,

The falling stone has covered a distance given by (1) as s = 4.9 × 4² = 78.4 m

Therefore, the stones will meet after 4 s at a height (100 – 78.4) = 20.6 m from the ground.