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Water question

  1. Sep 12, 2006 #1
    Suppose you adjust your garden hose nozzle for a hard stream of water. You point the nozzle vertically upward at a height of 1.5m above the ground. When you quickly move away from the vertical you hear the water striking the ground next to you for another 2.0s. What is the water speed as it leaves the nozzle?

    What I know/don't know:

    t (overall Time) = 2.0
    Total distance = d2 +d1 +1.5m and that d2 = d1 +1.5m


    Event 1
    Initial velocity = ?
    Final velocity = 0 m/s
    a= -9.80 m/s^2

    Event 2
    Initial velocity = 0 m/s
    Final velocity = ?
    a= -9.80 m/s^2


    I'm not sure how to proceed. I suspect I need a subsitution - may be time and distance? Any help is appreciated!
     
  2. jcsd
  3. Sep 12, 2006 #2

    HallsofIvy

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    What do d1 and d2 mean?

    No, the final velocity is NOT 0 m/s!

    event 2? What 2 events are you talking about?

    I assume you mean you hear the water striking the ground 2.0 seconds after starting the water rather than "for another 2.0s".

    The height, after t seconds, is given by x(t)= 1.5+ v0t- 4.9t2 (the 1.5 is the initial height above the ground, v0 is the initial velocity that you want to find and -4.9 is g/2). Saying that the water strikes the ground after 2 s means that x(2)= 1.5+ v0(2)- 4.9(22)= 0. Solve that for v0.
     
  4. Sep 12, 2006 #3
    I broke down d1 and d2 as distance 1 and distance 2. It looks like I didn't need to do that.

    I think you assumption is correct. My head is still spinning a bit on x(t)= 1.5+ v0t- 4.9t2 . I'm just wondering why it wasn't broken into two parts (the first part being the upward motion of the water to the arc and then the arc of the water to the ground).

    BTW thanks for help me with this problem. I appreciate it!
     
  5. Aug 30, 2008 #4
    so is the answer -18.1 m/s?
     
  6. Aug 30, 2008 #5

    HallsofIvy

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    How did you get that?
     
  7. Aug 30, 2008 #6
    this is how i got it

    x(t)= 1.5m+ 1/2(-9.8)(t)^2

    x(t)= 1.5-4.9(2)^2

    x(t)= 1.5-4.9(4)

    X(t)=-18.1

    *if this is wrong please show me the correct steps and explanation of what i did wrong please
     
  8. Aug 30, 2008 #7

    Redbelly98

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    The initial velocity should be part of that equation.
     
  9. Aug 31, 2008 #8
    i don't understand please show me
     
  10. Aug 31, 2008 #9

    Redbelly98

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    Your physics textbook has 3 or 4 standard equations that deal with constant-acceleration motion. You were using one of them when you wrote (post #6)

    Look in your book at the general equation where this comes from ... write out the equation as it's written in your book and show us what that equation is.
     
  11. Aug 31, 2008 #10
    the equation my book has is this X=Xo+Vot+1/2at^2 you put X(t) and that what im confused about shouldn't it be 1.5m=V(2)+1/2(9.8)(2)^2 and the final answer i got is -9.05m/s^2
     
  12. Aug 31, 2008 #11

    Redbelly98

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    Don't be confused, you are doing it correctly now :smile: By the way, it seems that you are defining upward as positive, and downward as negative.

    You correctly included Vo here, you did not include it in post #'s 4 and 6 (when you were getting a different answer).
     
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