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.