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14 November, 18:44

On another planet, a marble is released from rest at the top of a high cliff. It falls 4.00 m in the first 1 s of its motion. Through what additional distance does it fall in the next 1 s

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  1. 14 November, 20:44
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    Distance (h1) = 12 m

    Explanation:

    Given:

    Height of cliff (h) = 4m

    Time (t) = 1s

    Intial velocity (u) = 0m/s

    Computation:

    Distance (h) = 1/2 (g) (t) ²

    4 = 1/2 (g)

    g = 8m/s ²

    Distance (h1) = 1/2 (g) [ (t1) ² - (t) ²]

    Distance (h1) = 1/2 (8) (3)

    Distance (h1) = 12 m
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