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12 November, 11:37

A pair of glasses is dropped from the top of a 42.0 meter high stadium. A pen is dropped from the same position 2.00 seconds later. How high above the ground is the pen when the glasses hit the ground? (Disregard air resistance. a = - g = - 9.81 m/s ^2)

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  1. 12 November, 12:02
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    Given that,

    For glasses:

    distance, s = 42 m

    a = - g = - 9.81 m/s²

    we can calculate time taken by glasses to hit the ground as follow:

    s = Vi * t + 1/2 at²

    Since at top initial velocity, Vi = 0

    42 = 0 - 0.5 * 9.81 t²

    t = 2.926 s

    Pen is drop 2 second after the glasses is dropped. So time difference will be:

    Δt = 2.926 - 2 = 0.926 s

    Distance covered by Pen in 0.926 s can be calculated as:

    s = 0 - 0.5 * 9.81 * 0.926²

    s = 4.2 m

    Distance of pen from ground will be, 42-4.2 = 37.8 m

    When the glasses hit the ground, the pen will be 37.8 m above the ground.
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