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9 January, 07:07

A person stands on top of a tall building holding two rocks at a height of 50 meters. At the same moment, one rock is dropped from rest while the other is thrown downwards with an initial velocity of 8.0 m/s. Neglecting air resistance, calculate how long after the first rock lands does the second rock strike the ground

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  1. 9 January, 08:08
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    Answer

    The second rock will land 2.4s after the first rock

    Explanation:

    Given that

    Height of the building s=50m

    We assume that the first rock is acting with gravity so that a=9.81m/s

    And initial velocity u=0

    Applying the equation of motion

    S=ut+1/2at²

    50=0*t+1/2 (9.81) t²

    50=4.905t²

    t²=50/4.905

    t²=10.19

    t=√10.19

    t=3.19sec

    For the second rock initial velocity u=8m/s and v=0 and a=9.81

    Applying the equation of motion

    v=u+at

    0=8+9.81t

    t=-8/9.81

    t=0.81sec

    Hence the second rock will land 2.4s after the first rock

    I. e

    3.19-0.81
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