Physics

Gravitation

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 us assume both stones meet at time 't'.
Height of the tower (h) = 100m
Stone 1 dropped from the top of a tower:
initial velocity of the stone (u) = 0 m/s
acceleration due to gravity (g) is +ve = +9.8 m/s2
At time t, it will cover distance (h1), then using equation S = ut + ½at2
⇒ h1 = 0 + ½gt2 = (½)(9.8)t2 = 4.9t2



Stone 2 thrown upwards:
initial velocity (u) = 25 m/s
acceleration due to gravity (g) is negative = - 9.8 m/s2
At time t, 2nd stone reaches height (h2), then using equation S = ut + ½at2
⇒ h2 = 25t - ½gt2 = 25t - (½)(9.8)t2 = 25t - 4.9t2


Because Total height h = 100 m = h1 + h2
4.9t2 + 25t - 4.9t2 = 100
⇒ 25t - 100
⇒ t = 4s.
In 4 seconds, both stones will meet.
h1 (height from top) = 4.9 × 16 = 78.4m or 100- 78.4 = 21.6m from ground.
Both stones will meet at 21.6 m from ground.
(Note: If you take g = 10 m/s2 , the height will come as 20m)




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Gravitation

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Q 2.

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Q 3.

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Q 4.

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Q 5.

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Q 7.

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Q 8.

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Q 9.

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Q 10.

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Q 11.

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Q 12.

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Q 14.

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Q 19.

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Q 20.

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Q 21.

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Q 22.

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Q 24.

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Q 26.

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Q 27.

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Q 28.

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Q 29.

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Q 30.

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