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19 August, 08:07

Using energy considerations and assuming negligible air resistance, show that a rock thrown from a bridge 20.0 m above water with an initial speed of 15.0 m/s strikes the water with a speed of 24.8 m/s independent of the direction thrown. (Hint: show that $$K_i + U_i = K_f + U_f) $$

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  1. 19 August, 08:30
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    Given that,

    Height of the bridge is 20m

    Initial before he throws the rock

    The height is hi = 20 m

    Then, final height hitting the water

    hf = 0 m

    Initial speed the rock is throw

    Vi = 15m/s

    The final speed at which the rock hits the water

    Vf = 24.8 m/s

    Using conservation of energy given by the question hint

    Ki + Ui = Kf + Uf

    Where

    Ki is initial kinetic energy

    Ui is initial potential energy

    Kf is final kinetic energy

    Uf is final potential energy

    Then,

    Ki + Ui = Kf + Uf

    Where

    Ei = Ki + Ui

    Where Ei is initial energy

    Ei = ½mVi² + m•g•hi

    Ei = ½m * 15² + m * 9.8 * 20

    Ei = 112.5m + 196m

    Ei = 308.5m J

    Now,

    Ef = Kf + Uf

    Ef = ½mVf² + m•g•hf

    Ef = ½m * 24.8² + m * 9.8 * 0

    Ef = 307.52m + 0

    Ef = 307.52m J

    Since Ef ≈ Ei, then the rock thrown from the tip of a bridge is independent of the direction of throw
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