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Help with a hydrodynamics problem

  1. Oct 8, 2003 #1
    Ok, I edited the post to show all my work, so here it is...

    A hole is punched at a height h in the side of a container of height H. The container is full of water. If the water is to shoot as far as possible horizontally, (a) how far from the bottom of the container should the hold be punched? (b) Neglecting frictional losses, how far (initially) from the side of the container will the water land?

    Here's what I have so far:

    Using Bernoulli's Equation: P + .5*ρ*V^2 + ρ*g*h = a constant.
    and since flow rate(Q) = A*V = constant, I set the equations at the opening of the tank = to the hole that's been punched which gives:

    P1 + .5*ρ*V1^2 + ρ*g*h1 = P2 + V2^2 + ρ*g*h2

    P1 and P2 are 0 since the pressures are atmospheric. .5*ρ*V1^2 is assumed as 0, since the cross sectional area of the tank is much greater than the cross sectional area of the punched hole. I took the initial point at the waterline on top of the tank so ρ*g*h1 is 0 which leaves me with:

    .5*ρ*V2^2 = -ρ*g*h2 which simplifies to:

    V2 = √[2*g*(H - h)] which is the velocity of the water just as it exists through the punched hole in the tank.

    Using the kinematic equation:

    d = V2*t + .5*a*t^2

    t = √[(2*h)/g]

    Here's where I'm stuck. I've tried V2 * t = d and using first order derivatives to determine the maximum...but I get exactly twice the answer in the back of the book which is (H/2). I've posted this problem on another board and a member suggested using the quadratic formula but I couldn't really follow the person's reasoning so I came to the Physics Forums for different suggestions

    I'm sure (b) will be easy to find once I can figure out a way to solve for (a). Thanks.
     
    Last edited: Oct 8, 2003
  2. jcsd
  3. Oct 10, 2003 #2
    MisterNi,
    I think it's all OK until it comes to the kinematic equation.
    Your equation is for the water shooting up, not sideways.
    For shooting sideways, it is:
    h = g/2 t2 and
    d = v2t.
    This yields
    t = [squ](2h/g) and
    t = d/v2
    Eliminate t:
    [squ](2h/g) = d/v2
    Solve for d, substitute v2 with what you got, differentiate WRT to h, and find the maximum.
     
    Last edited: Oct 10, 2003
  4. Oct 12, 2003 #3
    Ah, thank you arcnets. The answer was staring me right in the face the entire time and I couldn't see it! Thanks again arcnets.

    And off topic, how do you make super and subscripted numbers?
     
  5. Oct 13, 2003 #4
    You type

    t [ sup ] 2 [ / sup ]

    or

    v [ sub ] 2 [ / sub ]

    without the blanks to get t2, v2
     
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