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6 January, 07:01

A 31.7 kg kid, initially at rest, slides

down a frictionless water slide at 53.2º.

How fast is she moving 3.45 s later?

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Answers (1)
  1. 6 January, 07:23
    0
    34.55m/s

    Explanation:

    Now from Newton's law of motion;

    V = U + a*t

    V - final velocity

    U - initial velocity = 0m/s

    a - acceleration

    t - time

    Note that a = g/cos53.2°

    Now g is acceleration of free fall due to gravity

    This is a vertical acceleration since the course of motion is along a plane inclined at 53.2° to the height of the pool.

    Also from trigonometry identity

    cos 53.2° = vertical component a / acceleration (cos a = adjacent/hypothenus)

    If we assume g = 9.8m/S2 { a generally given value}

    Substituting t = 3.45s as the time of consideration, we have;

    V = 0 + (9.8/cos53.2°) * 3.45

    = - 34.55m/s

    Note the - ve sign is just telling us this velocity is causing the object to move down.

    Hence V = 34.55m/s
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