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March 25, 2017

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A ball is thrown upward from a platform 4.5 m high with a speed of 17 m/s at an angle of 41 o from the horizontal. What is the magnitude of its velocity when it hits the ground?

  • physics - ,

    Vo = 17m/s @ 41deg.

    Vo(h) = 17cos41 = 12.8m/s.

    Vo(v) = 17sin41 = 11.15m/s.

    Vf^2 = Vo^2 + 2gd = 0.
    (11.15)^2 + (-19.6)d = 0,
    124.32 - 19.6d = 0,
    -19.6d = -124.32,
    d2 = 6.34m = distance above the platform.

    d = d1 + d2,
    d = 4.5 + 6.34 = 10.84m = Distance above ground.

    Vf^2 = Vo^2 + 2gd,
    Vf^2 = 0^2 + 2*(-9.8)*10.84 = 212.52,
    Vf = 14.6m/s = Final velocity.

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