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Hose Physics

  1. Jun 25, 2003 #1
    [SOLVED] Hose Physics

    Suppose you adjust your garden hose nozzle for a hard stream of water. You point the nozzle vertically upward at a height of 1.5 m above the ground (the hose is 1.5 m off the ground). When you quickly move the nozzle from the vertical you hear the water striking the ground next to you for another 2.0 s. What is the water speed as it leaves the nozzle?
     
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  3. Jun 25, 2003 #2

    Tom Mattson

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    Hi mt,

    I don't know if you noticed the Sticky I put at the top of this forum ("Read This Before Posting"), but it states the policy here that we want to see how you started and where you got stuck, then we help you through the rough spots.

    Thanks,
     
  4. Jun 25, 2003 #3
    Didn't notice that, but I left my work on a different thread, forgot to put it here...

    This how I got my answer, not sure if it was right though...

    v = V0T + 2AT^2 (2nd Kinematic equation)

    x-1.5 = v0(2s) + 2(9.8m/s^2)(2s)

    When T = 2, then x- 1.5 is zero (When the water hits the gound the velocity of the water is zero)

    v0 = -19.6m/s or 19.6 m/s
     
  5. Jun 25, 2003 #4
    Forgot to divide the 2s, its 9.8 m/s not 19.6
     
  6. Jun 25, 2003 #5

    Tom Mattson

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    Not quite. That should be:

    x=x0+v0t+(1/2}at2

    Just change the 2 to (1/2)[color] and recalculate.

    Yes.

    No. The speed of the water is only zero at the top of the trajectory. By the time it hits the ground, it will be moving even faster than it was when it left the nozzle.

    You'll have to correct the mistakes noted above and try again.
     
  7. Jun 25, 2003 #6
    v = V0T + .5AT^2

    0 m/s = V0(.30s) + .5(9.80 m/s^2)(2s)^2

    V = -65.3 m/s or 65 m/s

    This doesn't seem right, a garden house shooting 65+ feet in the air! It must have something to do with the 0 for the final velocity (the water resting on the ground). Other than that I really can't see what I am doing wrong.
     
  8. Jun 25, 2003 #7

    Tom Mattson

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    You didn't correct the first mistake I pointed out.

    The equation should be:

    x(not v!)=x0+v0t+(1/2)at2
     
  9. Jun 25, 2003 #8
    Sry, typo, its hot out here :), despite that fact I made the x equal x-1.5. When you set T = 2, x - 1.5 = 0 because when the water hits the ground it no longers has a velocity.
     
  10. Jun 25, 2003 #9

    Tom Mattson

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    Yeah, it's hot here too. :frown:

    Whoa!

    First, I am supposing that you set x=0 at ground level, so x-1.5 would indeed equal zero. But that has nothing to do with the velocity of the water.

    Second, it is not true that the water does not have a velocity when it hits the ground. Indeed, its speed is greater when it hits the ground than it was when it left the nozzle. I already said that in my first post.

    The equation for velocity is:

    v=v0+at

    In this case, a=-g (I take g to be a positive number), and you can see for yourself that v is definitely not zero when the water hits the ground. To find that out, you have to solve for t in the equation for x, then plug t into the equation for v.
     
  11. Jun 25, 2003 #10
    But we aren't solving for V, we are solving for initial velocity. Moreover, how can I solve t in the equation for x if I dont the initial velocity:

    x = x0 + v0t +.5at^2

    X = 0
    x0 = 1.5 m
    T = unknown
    A = 9.8 m/s^2
    V0 = unknown

    Two variables? Something has to be done with the piece of information telling us that the water continues to hit the ground for 2 minutes, I just dont know what.
     
  12. Jun 25, 2003 #11
    Correction: 2 seconds
     
  13. Jun 25, 2003 #12

    Tom Mattson

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    I know that. I told you that you can solve for v to convince yourself that the speed of the water is not zero when it hits the ground. I said that because you told me twice that it is zero, and that is wrong.

    But you were given T. They tell you that you water continues to land next to you for 2 seconds. That means it took 2 second for the water to go up and then come back down.
     
  14. Jun 25, 2003 #13
    x = x0 + v0t +.5at^2

    X = 0
    x0 = 1.5 m
    T = 2 secs
    A = 9.8 m/s^2
    V0 = unknown

    0 m = 1.5 m + v0(2s) + .5(9.80m/s^2)(4 s) =

    0 m = 1.5 m + v0(2s) + 19.6 m/s^3 =

    0 m = 21.1m/s^3 + v0(2s) =

    -21.1m/s^3 = v0(2s) =

    -10.55 m/s^2 = v0

    Isn't this a form of acceleration rather than velocity?
     
  15. Jun 25, 2003 #14

    Tom Mattson

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    No, you just made a mistake with the t2 term.

    (2s)2=4s2, but you have 4s.
     
  16. Jun 25, 2003 #15
    Thank you very much for your helf throughout the day; it is much apprectiated.
     
  17. Jun 25, 2003 #16
    Thank you very much for your helf throughout the day; it is much apprectiated.
     
  18. Jun 26, 2003 #17
    Hi, last time I looked at this, there were no replies. In the meantime, I found a solution. I think it's safe to post it now, since so much discussion has been gong on...

    In a free fall from height h, the time is
    t = sqrt(2h/g).
    In this problem, we have a rise of h, and then a fall of h+s (where s = 1.5m). So,
    t = sqrt(2h/g) + sqrt(2(h+s)/g).
    Now, if we substitute h = v^2/2g, we get
    t = v/g + sqrt((v/g)^2 + 2s/g).

    It's quite easy to solve this for v.
     
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