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Bernoulli's equation and a water tank

  1. Nov 5, 2008 #1
    1. The problem statement, all variables and given/known data

    A water tank of height h has a small hole at height y. The water is replenished to keep h from changing. The water squirting from the hole has range x. The range approaches zero as y goes to 0 because the water squirts right onto the table. The range also approaches zero as y goes to h because the horizontal velocity becomes zero. Thus there must be some height y between 0 and h for which the range is a maximum.

    a)Find an algebraic expression for the flow speed v with which the water exits the hole at height y.

    b)Find an algebraic expression for the range of a particle shot horizontally from height y with speed v.

    c)Combine your expressions from parts A and B. Find the maximum range x_max

    d)Find the height y of the hole.

    2. Relevant equations

    v1A1=v2A2
    p1+.5(density)v1^2+(density)gy1=p2+.5(density)v2^2+(density)gy2
    deltaK+deltaU=Wext


    3. The attempt at a solution

    a)Solving for v2(assuming this is outside the can velocity):

    v=sqrt of 2p1/denisty-2p2/density+v1^2+2gy1-2gy2

    computer told me to check my answer, so I guess its wrong.

    b)distance=rate*time

    t=sqrt of 2h/g

    x=v*t

    haven't tried part c or d cause I cant get part a) right.

    Thanks for any help.
     
    Last edited: Nov 5, 2008
  2. jcsd
  3. Nov 5, 2008 #2
    Ok, so I have messed around a bunch and I think the answer for part a) is:

    v=sqrt of 2*g(h-y)

    Can anyone tell me if I am on the right track?
     
  4. Dec 1, 2008 #3
    that's what i got for v, what did u get for the max range of x??
     
  5. Dec 2, 2008 #4
    I typed in your equation for v and its correct. For my services, you think you could explain a bit how you got it?
     
  6. Dec 2, 2008 #5
    Oh just realized that was posted some time, probably doesnt need my services, but . . .

    Bluebear, I'm stuck at xmax = sqrt(4y(h-y)) I literally just substituted the answer from A into B.

    But I think we have to get rid of y somehow so I'm hesitant to submit this. Have you gotten it yet?
     
  7. Dec 2, 2008 #6
    the velociyty for an obj falling in kinematics is sqrt[2gh]
    Vf^2 - Vi^2 = 2as where a = acceleration and s = displacement
    here a = g since its falling and s = h-y = distance the water has fallen up to the point of the hole.
    and Vi = 0
    so Vf^2 = 2*g*(h-y)
    and solve for v by taking the square root of both sides
    and no I havent tried c yet but it is part of my homework so I will try very soon
    hope this helps
     
  8. Dec 2, 2008 #7
    the eqn you got i think only solves for x, but not Xmax

    y = 1/2 h for max range, i think you can solve for the rest
    you had the right idea with sqrt(4y(h-y)) take the 4 out and u get 2sqrt(y(h-y))
    looking at inside the radical y(h-y). you want that value to be the largest as possible. But when you increase y, u decrease h-y and vice versa, so the biggest value u can get is 1/2 * 1/2. its a little common sense twist
     
    Last edited: Dec 2, 2008
  9. Dec 3, 2008 #8
    Great, thanks for your help. You havent by any chance also been assigned 15.58, the boat problem? There's a thread called "buoyancy force on a steel boat", where I describe my work on it.
     
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