# A stone is allowed to fall from the top of a tower 100 m high

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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.

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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.

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