Question

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. 

Answer 1

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, 

= + 1/ 2 ² 

= 0 × + 1/ 2 × 9.8 × ² 

⇒ = 4.9 ² … … … … … … … … . (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 ′. 

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

Equation of motion, 

= + 1 /2 2 

′ = 25 × − 1/ 2 × 9.8 × ² 

⇒ ′ = 25 − 4.9 ² … … … … … … … … . (2) 

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

′ + = 100 

⇒ 25 − 4.9 2 + 4.9 ² = 100 

⇒ = 100/ 25 = 4 

In 4 s, 

The falling stone has covered a distance given by (1) as = 4.9 × 4 ² = 78.4  

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