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16 April, 06:18

A ski lift is used to transport people from the base of a hill to the top. If the lift leaves the base at a velocity of 15.5 m/s and arrives at the top with a final velocity of 0 m/s, what is the height of the hill? Round the answer to the nearest tenth

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  1. 16 April, 06:26
    0
    This can be calculated with the law of conservation of energy. The sky lift is starting with the speed v = 15.5 m/s and all of it's kinetic energy Ek is transformed to potential energy Ep so the energies have to be equal: Ep=Ek.

    Since Ek = (1/2) * m*v² where m is mass and v is the speed, Ep=m*g*h, where m is mass, g = 9.81 m/s² and h is height. Now:

    Ek=Ep

    (1/2) * m*v²=m*g*h, masses cancel out,

    (1/2) * v²=g*h, divide by g to get the height,

    (1/2*g) * v²=h and now plug in the numbers:

    h=12.245 m. Height of the hill rounded to the nearest tenth is h=12.25 m
  2. 16 April, 06:30
    0
    Just to keep it simple for anyone looking for the answer while they're on Edge, just type in 12.3
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