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Water being hosed into the air.

  1. Sep 7, 2013 #1
    1. The problem statement, all variables and given/known data

    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 the nozzle 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?

    2. Relevant equations

    Δx =Vo t + 1/2 at^2

    3. The attempt at a solution

    Δx =Vo t + 1/2 at^2
    -1.5m = (2s) Vo + (1/2) (-9.8 m/s^2) (2s)^2
    -1.5m = (2s) Vo - 19.6 m
    Vo = 9.05 m/s

    Now the above solution seems to be correct, but I am not sure why t should = 2s.
    I ended plugging 2 sec. in without really thinking about it, but now I am not able to
    spatially visualize what is going on with the water.

    Any help appreciated.
     
  2. jcsd
  3. Sep 7, 2013 #2
    This is just a very standard linear kinematics question that's been verbally tarted up. The point is that once the hose is moved away from vertical the last 'drop' of water to leave the nozzle hits the ground 2s later. So you're right to use 2s as the time in this case. This problem is mathematically identical to the usual rock/ball/textbook/whatever being thrown into the air.
     
  4. Sep 7, 2013 #3
    Hello!

    Thank you, that makes sense now. :)
     
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