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a stone is allowed to fall from the top of a tower...

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:

Let the two stones meet after a time t.

(i) For the stone dropped from the tower:

Initial velocity, u = 0 m/s

Let the displacement = s

Acceleration due to gravity, g = 9.8 m s⁻²

From the equation of motion,

...(1)

(ii) For the stone thrown upwards:

Initial velocity, u = 25 m/s

Let the displacement = s'.

Acceleration due to gravity, g = −9.8 m s⁻²

Equation of motion,

...(2)

The combined displacement is ;

In 4 s, the falling stone has covered a distance given by equation (1) as

Therefore, the stones will meet after 4 s and the distance is 80 m from the top.

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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.

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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.

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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.