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Torricelli's Law Problem

  1. Apr 21, 2015 #1
    1. The problem statement, all variables and given/known data
    (This is a continuation of the problem where I proved Torricelli's Law: v = (2gh))

    The water level in a tank lies a distance H above the floor. There is a hole in the tank that a distance h below the water level
    a.) Find the distance x from the wall of the tank at which the leaked stream of water hits the floor
    b.) Could another hole be punched at another depth h' so that this second stream would have the same range? If so, at what depth?

    2. Relevant equations
    Δx = (vf2 - vi2)/g
    v = (2gh)

    3. The attempt at a solution
    At first, I was not really sure what to do here at all (and I probably ended up solving it completely wrong.)

    So, basically, I wasn't exactly sure how to find the distance, since I only knew the initial velocity in the x direction (or, at least, I assumed v = (2gh)) was the velocity of the water in the x direction based off the picture in the problem.)

    So, what I tried was:
    (Since the equation I had to derive from Torricelli's law in the problem I had to do before this ended up being Δv = (2gh), I assumed:)

    Δx = (2gh)/g
    = 2h

    For b, I'm completely confused on even where to begin solving for the equation.
     

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  3. Apr 21, 2015 #2

    mfb

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    Where does that Δx formula come from?
    Your water, once it leaves the tank, moves horizontally and is in free fall. What do you know about objects in free fall?
     
  4. Apr 21, 2015 #3
    Wait a minute...the water has a vy when it leaves the tank of 0 m/s, doesn't it? If so, then I think I can solve it.
     
  5. Apr 21, 2015 #4

    mfb

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    Right, assuming "y" is your vertical coordinate.
     
  6. Apr 21, 2015 #5
    Well, I over-thought that problem.

    Then:

    vi-y = 0 m/s
    So:
    ==> vf-y = vi-y + 2gh
    = 0 + 2gh
    = 2gh

    So:
    vf-y = vi-y + gt
    t = vf-y/g
    ==> t = (2gh)/g

    So:
    This is where I use: v = (2gh)
    Δx = vi-yt + 0.5at2
    = (2g*(H-h)) * (2gh)/g
    = 2 √(h(H-h))

    Still not sure about b, though. Would I just work backwards?
     
  7. Apr 21, 2015 #6

    haruspex

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    If a hole at height h' produces the same Δx, what equation can you write?
     
  8. Apr 22, 2015 #7

    mfb

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    A velocity cannot be equal to 2gh, the units do not match. And I don't understand the other steps, but the result looks right.
     
  9. Apr 22, 2015 #8
    v = (2gh) was what I had to prove in the previous problem.
     
  10. Apr 22, 2015 #9
    Wait, I see the issue. In my notes, I think I had a square root sign. It just never made it onto paper.
     
  11. Apr 22, 2015 #10
    Well, if it's at h', then the Vi-x will be √(2g(H-h')) and you can work backwards from there.
     
  12. Apr 22, 2015 #11

    haruspex

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    Better, consider h' in respect of this equation in your post #5: Δx = 2 √(h(H-h)).
     
  13. Apr 23, 2015 #12

    mfb

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    Drawing a sketch of Δx as function of h could be useful.
     
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