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.