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Abody falling freely under gravity passed two points 30 m apart in 1s from what point above the upper point it began to fall

Answers

  • Réponse publiée par: lhadyclaire

    Step-by-step explanation:

    firstly we have to find the v

    so

    v=u+at

    v= 0+(9.8)(1)

    v=9.8m/s

    so for distance

    2as=v^2 -u^2

    2(9.8)(s) = 9.8^2

    2×9.8(s) = 9.8×9.8

    s= 9.8/2

    s= 4.9 metre

  • Réponse publiée par: RoseTheShadowHunter

    S = v t + 1/2 g t^2      where S = 30 m   and t = 1 s

    30 = v * 1 + (9.8 / 2) * 1

    v = 30 - 4.9 = 25.1

    So V = 25.1 m/s   when it passed the upper point

    V = g T   where T is the time elapsed to reach the upper point

    T = 25.1 / 9.8 = 2.56 s

    H = 1/2 g t^2 = (9.8 / 2) * 2.56^2    where H is height above upper point

    H = 32.1 m

    As a check:

    Vu = 9.8 * 2.56 = 25.1 m/s    speed at upper point

    Vl = 9.8 * 3.56 = 34.9 m/s     speed at lower point

    2 g S = 34.9^2 - 25.1^2 = 588

    S = 588 / 19.6 = 30 m

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Abody falling freely under gravity passed two points 30 m apart in 1s from what point above the uppe...