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Couple of pretty simple liquid q's

  1. Dec 9, 2004 #1
    1) First ones just im curious about whats the relationship between the velocity of a liquid being projected, like out of a hose, vertically and the height it goes.

    2) An actual problem: A cylindrical tank .9 m in radius rests atop a platform 6 m high. Initially the tank is filled with water to a depth of h=3m. A plug whose area is 6.3 cm^2 is removed from an orifice in the side of the tank at the bottom. What is the speed of the stream as it strikes the ground.

    Will the water have an initially speed when it comes out? Obviously gravity is what is going to accelerate it down at 9.8 m/s, I thought that it would just be the acceleration multiplied by the time because it wouldnt have an initial speed, it was just coming out. This seems to simple and doesnt use the informatyion they gave me so guessing its not the answer
  2. jcsd
  3. Dec 9, 2004 #2


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    It isn't simple.It's not too difficult,either...Well,...to me,that is...
    Calculate the speed at which the particles (molecules if u prefer) at the surface of the water inside the tank begin to fall,once u've opened the plug.Use Bernoulli's equation of continuity for incompressible fluids (the one that states that the flow is conservative),to find at what speed the particles of water exit the tank through the hole.
    That would be your initial speed of the water that comes out of the tank.Plug it in the equation for motion in the gravitational field to find the final speed (the speed at the contact with the ground).

  4. Dec 9, 2004 #3
    P1 + 1/2PV^2 + PGY1 = P2 + 1/2PV2^2 + PGY2

    (I know the P's in 1/2PV^2 are rho for density, new here with the greek letters)

    P1 and P2 both 1 atm?
    Y1 = The top of the barrel so 3m since its filled up 3m?
    Y2 = 0?

    If that is the correct way I should be all set
  5. Dec 9, 2004 #4
    Or should Y1 be 9 and Y2 be 3 since the barrel is 6m above the ground
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