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A novice skier, starting from rest, slides down a frictionless 33.0° incline whose vertical height is 127 m. How fast is she going when she reaches the bottom?

  • physics -

    The inclined plane forms a rt. triangle:

    Y(ver) = 127 m,
    A = 33 deg = angle bet. hor side and hyp.

    d(hyp) = Y / sinA = 127 / sin33 = 233.2 m.

    Vf^2 = Vo^2 + 2gd,
    Vf^2 = 0 + 2 * 9.8 * 233.2,
    Vf^2 = 4570.7,
    Vf = 67.6 m/s = Final velocity.

  • physics -

    V=sq.root(2gh)=sq.root-- 2 x 9.8m/s x 127m=49.89m/s

    Angle does not really matter.

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