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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.- Physics Cbse class 11

Asked On2017-07-24 08:50:11 by:milan-ransingh

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Download question setAnswersLet the two stones meet after a time t.

(i) For the stone dropped from the top of the tower:

Initial velocity, u = 0

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,

(ii) For 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

From the equation of motion,

The combined displacement (s + s') of both the stones at the meeting point is equal to the height of the tower 100 m.

From eqs. (1) and (2), we get,

s + s' = 4.9 t

^{2}+ 25 t - 4.9 t^{2}100 = 25 t

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

s = 4.9 t

^{2}= 4.9 × (4)^{2}= 78.4 mTherefore, the stones will meet after 4 s at a height (100 - 78.4) = 21.6 m from the ground.

Concept Note:Choose the equation of motion wisely to minimize the number of steps in calculations.Likes:

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