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Liquid Problem

  1. Dec 12, 2004 #1
    A Hose shoots water straight up for a distance of 2.5m. The end opening on the hose has an area of .75cm^2. (A) What is the speed of the water as it leaves the hose? (B) How much water comes out in 1 minute.

    If I have A i can find B, I just can't figure out how to find A. I know gravity is going to be what is pulling it down, I just am unsure of what equation to use. I was thinking maybe Bernouli's Equation but not positive, any help is appreciated
     
  2. jcsd
  3. Dec 12, 2004 #2

    Doc Al

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    Staff: Mentor

    Treat the water as any other "projectile". What speed does a projectile need to rise up that distance?
     
  4. Dec 12, 2004 #3
    Sorry for lack of symbols, still figuring it out how to work them all but..

    Y = Y0 + V0T + -1/2GT^2

    Y0 = Original Height
    V0 = Original Velocity

    So..

    2.5 = V0 - .5(9.8)

    So 2.5 + 4.9 which is 7.4?

    I may of mixed up the equation, forgot book in my locker which is part of the problem
     
  5. Dec 12, 2004 #4

    Doc Al

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    That equation relates distance and time. But you aren't given the time.

    You need a kinematic equation that relates distance and speed.
     
  6. Dec 12, 2004 #5
    V^2 = V0^2 - 2aY or
    V^2 = V0^2 + 2a(Y1 - Y0)

    The water at its very peak V will equal 0 won't it? So..

    0 = V0^2 - 2(9.8)(2.5 - 0)
    0 = V0^2 - 49
    V0^2 = 49
    V0 = 7
     
  7. Dec 12, 2004 #6

    Doc Al

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    Right! But don't forget units: 7 m/s.
     
  8. Dec 12, 2004 #7
    Thanks a lot, i really appreciate it
     
  9. Dec 12, 2004 #8
    A comment...I'd do that problem by considering a small "piece" of the fluid, finding its potential energy at its maximum height (m*g*h), and equating that to its initial kinetic energy.
     
  10. Dec 13, 2004 #9

    Doc Al

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    An excellent approach which leads to the exact same equation:
    [itex]mgh = 1/2mv^2[/itex] => [itex]v^2 = 2gh[/itex]
     
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