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24 October, 22:48

a stone is thrown by a person from the top of the building, which is 200m tall. at the same time, another stone is thrown with verlocity of 50m/s by a person standing at the foot of the building. find the time after which the two stones meet.

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  1. 25 October, 01:22
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    The time after which the two stones meet is tₓ = 4 s

    Explanation:

    Given data,

    The height of the building, h = 200 m

    The velocity of the stone thrown from foot of the building, U = 50 m/s

    Using the II equation of motion

    S = ut + ½ gt²

    Let tₓ be the time where the two stones meet and x be the distance covered from the top of the building

    The equation for the stone dropped from top of the building becomes

    x = 0 + ½ gtₓ²

    The equation for the stone thrown from the base becomes

    S - x = U tₓ - ½ gtₓ² (∵ the motion of the stone is in opposite direction)

    Adding these two equations,

    x + (S - x) = U tₓ

    S = U tₓ

    200 = 50 tₓ

    ∴ tₓ = 4 s

    Hence, the time after which the two stones meet is tₓ = 4 s
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