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Homework Help: Easy-ish Conservation of Energy question!

  1. Jul 19, 2010 #1
    1. The problem statement, all variables and given/known data
    A waterfall has a change in elevation of 4.4*10^2 m. When the water has fallen 12% of its way to the bottom, its speed is 93m/s. Neglecting air resistance and fluid friction, determine the speed of the water at the top of the waterfall.

    2. Relevant equations
    Et1=Et2
    Ek1+Eg1=Ek2+Eg2
    1/2m(v1)^2+mgh1=1/2m(v2)^2+mgh2

    3. The attempt at a solution
    g=9.81 N/kg
    v1=?
    v2=93m/s
    h1=440m
    h2=440m*0.88=387.2m

    1/2m(v1)^2+mgh1=1/2m(v2)^2+mgh2
    1/2(v1)^2+gh1=1/2(v2)^2+gh2
    1/2(v1)^2+(9.81)(440m)=1/2(93)^2+(9.81)(387.2m)

    Am I doing it right so far?
     
  2. jcsd
  3. Jul 19, 2010 #2

    Doc Al

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    Staff: Mentor

    So far, so good.
     
  4. Jul 19, 2010 #3
    1/2m(v1)^2+mgh1=1/2m(v2)^2+mgh2
    1/2(v1)^2+gh1=1/2(v2)^2+gh2
    1/2(v1)^2+(9.81)(440m)=1/2(93)^2+(9.81)(387.2m)
    1/2(v1)^2+(4316.4)=4324.5+3798.432
    1/2(v1)^2=3824.532
    v1=87.5m/s

    But this is not the answer in the answers section of the textbook.
     
  5. Jul 19, 2010 #4

    Doc Al

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    Staff: Mentor

    The answer I get is very close to yours. (Just differs in the third digit.)

    What textbook, by the way?
     
  6. Jul 19, 2010 #5
    Maybe the answer is wrong in the textbook, it says 5m/s.
    It's Nelson Physics 12 from 2003 I think.
     
  7. Jul 20, 2010 #6

    Doc Al

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    Staff: Mentor

    Sounds like the book messed up. If it started out at 5 m/s and fell the full 440 m, then it would have a speed of 93 m/s at the bottom.
     
  8. Jul 28, 2010 #7
    Silly textbook, thanks though! :)
     
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