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28 July, 13:31

Imagine that you are working as a roller coaster designer. You want to build a record breaking coaster that goes 70.0 m/s at the bottom of the first hill. You estimate that the efficiency of the tracks and cars you are using 90.0%. How high must the first hill be?

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  1. 28 July, 14:47
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    Wow! This is not simple. At first, it looks like there's not enough information, because we don't know the mass of the cars. But I"m pretty sure it turns out that we don't need to know it.

    At the top of the first hill, the car's potential energy is

    PE = (mass) x (gravity) x (height).

    At the bottom, the car's kinetic energy is

    KE = (1/2) (mass) (speed²).

    You said that the car's speed is 70 m/s at the bottom of the hill,

    and you also said that 10% of the energy will be lost on the way

    down. So now, here comes the big jump. Put a comment under

    my answer if you don't see where I got this equation:

    KE = 0.9 PE

    (1/2) (mass) (70 m/s) ² = (0.9) (mass) (gravity) (height)

    Divide each side by (mass):

    (0.5) (4900 m²/s²) = (0.9) (9.8 m/s²) (height)

    (There goes the mass. As long as the whole thing is 90% efficient,

    the solution will be the same for any number of cars, loaded with

    any number of passengers.)

    Divide each side by (0.9):

    (0.5/0.9) (4900 m²/s²) = (9.8 m/s²) (height)

    Divide each side by (9.8 m/s²):

    Height = (5/9) (4900 m²/s²) / (9.8 m/s²)

    = (5 x 4900 m²/s²) / (9 x 9.8 m/s²)

    = (24,500 / 88.2) (m²/s²) / (m/s²)

    = 277-7/9 meters

    (about 911 feet)
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