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Speed of water from a sprinkley

  1. Dec 10, 2008 #1
    1. The problem statement, all variables and given/known data
    Four lawn sprinkler heads are fed by a 1.9cm diameter pip.e The water comes out of the heads at angle of 35 degrees to the horizontal and covers a radius of 8m.
    a) What is the velocity of the water coming out of each sprinkler head?
    b) If the output diameter of each head is 3.0mm, how many liters of water do the four heads deliver per second?


    2. Relevant equations
    Continuity: A1*v1 = A2*v2


    3. The attempt at a solution
    For (a), I see the water covers a radius of 8m, so an area of 64pi meters. It must do this in a horizontal direction, so the speed is only in the X direction. I don't really know how to find out velocity though, because we don't know any other velocity.
    For (b) I assume once you just multiply the area by the velocity found in (a)
     
  2. jcsd
  3. Dec 10, 2008 #2

    Doc Al

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    Staff: Mentor

    Hint: When the water leaves the sprinkler head, it becomes a projectile. (What if, instead of water, it was a ball throw at a given angle? How would you solve it then?)
    Yes, the flow rate from each head is given by V*Area.
     
  4. Dec 10, 2008 #3
    I don't know what to do here. I know for a projectile you would figure out the total time in the air based on the y-component of velocity, then you can figure out the x component of velocity. So in this problem Vy = Vsin(35) and Vx=Vcos(35). How does this help though if I don't know V?
     
  5. Dec 10, 2008 #4

    Doc Al

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    You're asked to find V. You know the distance traveled--use it.
     
  6. Dec 10, 2008 #5
    Haha I don't know how to. There are 2 variables missing- V (which I have to find) and t. I would say get t from another equation, but I can't get it from the y=yo +Vyo + .5at^2 because y is missing.
     
  7. Dec 10, 2008 #6

    Doc Al

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    Luckily you have two equations.
    What's the initial and final y position?
     
  8. Dec 10, 2008 #7
    Is it .5at^2=Vsin35t, so t=Vsin(35)/4.9?

    If so, I get V=83.4 which seems really really fast.
     
  9. Dec 10, 2008 #8

    Doc Al

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    That's true.
    That's not. Redo that last bit.
     
  10. Dec 10, 2008 #9
    oh sqrt, so V= 9.13. Cool.

    Just one more quick question, pretty related. I did a simple problem earlier today where it was like "Romeo wants to throw a ball so it hits Juliet's windows with only the x component of velocity." They gave you the angle which he threw it at.
    In that problem they set Vyo=0 and solved for t based off the height provided. How could you set Vyo=though? If Vyo=0, wouldn't Vxo also =0?
    As for this case, the reason Vyo isn't =0 is because the water is moving inside the pipe and comes up and out.
     
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