Physics

A man skis down a slope 120 m high. If 80 percent of his initial potential energy is lost to friction and air resistence, what is his speed at the bottom of the slope?

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  1. 20% of M g H is converted to (1/2) M V^2
    Note that the mass M cancels out.

    0.2 gH = 0.5 V^2

    H = 120 m; g = 9.81 m/s^2

    Now just Solve for V

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  2. 45

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  3. PE=KE
    mgh=1/2mv^2
    0.2(9.8m/s^2)(120m)=0.5(v^2)
    (235.2m^2/s^2)/0.5=(0.5v^2)/0.5
    v^2=470.4
    v=21.689 m/s

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