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Need quick turnaround

  1. Nov 21, 2008 #1
    1. The problem statement, all variables and given/known data Instructor has not gone over this material and I want to know how to work this type of problem before she does. I don't even know how to begin.

    A swimming pool is filled with water. It is 2.50 m tall and 3.00 m in diameter. There is a small 1.00 cm (in diameter) hole in the side of the pool and its 0.50 m below the top. How far from the pool will the water coming out of the hole land?



    2. Relevant equations



    3. The attempt at a solution
    OK, here's my attempt using Torricelli's theorem, however I need to use Bernoulli's equation any help is appreciated.

    Vx = Sqrt 2g(2.50 - 1.00)
    Vx = 5.42 m/s This is the horizontal velocity.

    t= sqrt 2h/g = .20 seconds = time water is in air

    x = t x Vx

    x = 1.11 meters ????
     
    Last edited: Nov 21, 2008
  2. jcsd
  3. Nov 21, 2008 #2

    Doc Al

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    Staff: Mentor

    What have you tried so far?

    Hint: Look up Torricelli's theorem.
     
  4. Nov 21, 2008 #3

    mgb_phys

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    Bernoulli's equation will give you the sideways velocity of the water out of the hole.
    Then it's just the normal vertical acceleration equation to find the time before it hits the ground (like a cannon ball fired horizontally)
     
  5. Nov 21, 2008 #4
    Here's my attempt using Torricelli's theorem but I need to use Bernoulli's equation

    Vx = Sqrt 2g(2.50 - 1.00)
    Vx = 5.42 m/s This is the horizontal velocity.

    t= sqrt 2h/g = .20 seconds = time water is in air

    x = t x Vx

    x = 1.11 meters ????

    Doesn't the diameter of the pool and the size of the hole have any play in the equation?
     
  6. Nov 21, 2008 #5

    Doc Al

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    They are intimately related. (You can derive Torricelli's theorem from Bernoulli's theorem using certain simplifying assumptions.)

    Where did you get those heights?

    Again, what height are you using?


    Not much. Compare the speed of water at the top compared to that at the hole.
     
  7. Nov 21, 2008 #6
    Vx = Sqrt 2g(2.50 - 2.00)

    Vx = 3.13 m/s horiz vel

    3.13 x .41 = 1.28 m

    Thanks for pointing out my calc in heights this is what I came up with based on 2.50 m = h0 2.00 m = h

    How do I solve the same using Bernoulli's equation and is my answer of 1.28 m right?
     
    Last edited: Nov 21, 2008
  8. Nov 21, 2008 #7

    Doc Al

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    Looks good.

    Correct that value for time.

    Set up Bernoulli's equation to compare a point at the top of the pool to a point at the hole.
     
  9. Nov 21, 2008 #8
    t= sqrt 2 x 2/9.8 = .64 s

    so 3.13 x .64 = 2.0 m
     
  10. Nov 21, 2008 #9

    Doc Al

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    Looks good.
     
  11. Nov 21, 2008 #10
    Can you view my post again and determine if I came up with the same answer if I had used Bernoulli's equation and then calculated the time?
     
    Last edited: Nov 21, 2008
  12. Nov 22, 2008 #11

    Doc Al

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    If you apply Bernoulli's equation correctly, and make the same simplifying assumptions used in Torricelli's theorem, then you'll get the same answer. After all, that's where Torricelli's theorem comes from--it's just a special case of Bernoulli's equation.

    Again, I urge you to actually look up Torricelli's theorem (in your book or on the web) and see how it's derived.
     
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