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25 June, 07:09

A ball is thrown upward from the top of a 25.0 m tall building. The ball's initial speed is 12.0 m/sec. At the same instant, a person is running on the ground at a distance of 31.0 m from the building. What must be the average speed of the person if he is to catch the ball at the bottom of the building?

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  1. 25 June, 10:28
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    Person must have 8.18 m/s to catch the ball

    Explanation:

    Consider the vertical motion of ball

    We have equation of motion s = ut + 0.5at²

    Initial velocity, u = 12 m/s

    Acceleration, a = - 9.81 m/s²

    Displacement, s = - 25 m

    Substituting

    -25 = 12 x t + 0.5 x - 9.81 x t²

    4.905 t² - 12t - 25 = 0

    t = 3.79 sec

    Ball hits ground after 3.79 seconds.

    So person need to cover 31 m in 3.79 seconds

    Consider the horizontal motion of person

    We have equation of motion s = ut + 0.5at²

    Initial velocity, u = ?

    Acceleration, a = 0 m/s²

    Displacement, s = 31 m

    Time, t = 3.79 seconds

    Substituting

    31 = u x 3.79 + 0.5 x 0 x 3.71²

    u = 8.18 m/s

    Person must have 8.18 m/s to catch the ball
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