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12 June, 00:33

A 20 kg child is on a swing that hangs from 3.0-m-long chains. what is her maximum speed if she swings out to a 49 ∘ angle?

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  1. 12 June, 00:55
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    L = length of the chain = 3 m

    θ = angle with the vertical = 49 deg

    h = height gained

    In triangle ABD

    Cosθ = AB/AD

    Cosθ = AB/L

    AB = L Cosθ

    BC = AC - AB = L - L Cosθ

    so h = L - L Cosθ

    inserting the values

    h = (3) - (3) Cos49 = 1.032 m

    v = maximum speed at the bottom of the swing

    Using conservation of energy

    kinetic energy at the bottom of the swing = potential energy at the maximum height

    (0.5) m v² = mg h

    v = sqrt (2gh)

    v = sqrt (2 x 9.8 x 1.032)

    v = 4.5 m/s
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