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Projectile motion of a fire hose

  1. Jan 25, 2009 #1
    Firemen are shooting a stream of water at a burning building using a high-pressure hose that shoots out the water with a speed of 25.0 {\rm m/s} as it leaves the end of the hose. Once it leaves the hose, the water moves in projectile motion. The firemen adjust the angle of elevation \alpha of the hose until the water takes 3.00 {\rm s} to reach a building 45.0 {\rm m} away. You can ignore air resistance; assume that the end of the hose is at ground level.


    I found the following:
    angle a = 53.1
    speed of water at highest point = 15m/s
    magnitude of acceleration at highest point = 9.8m/s^2

    I need to find:
    How high above the ground does the water strike the building?
    How fast is it moving just before it hits the building?

    I tried using V^2 - Voy^2 = -2g(y-yinitial)

    I can't seem to get the right answer.
     
  2. jcsd
  3. Jan 25, 2009 #2
    I figured out how to solve for the height above the ground and got 15.9m which is correct.
    Now I need How fast is it moving just before it hits the building?
     
  4. Jan 25, 2009 #3

    LowlyPion

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    How fast it was going.

    That means they want the horizontal velocity and the vertical velocity at that point.

    The horizontal is easy. V*cosθ

    The vertical at that height isn't all that hard either, if you recognize that in a gravity field the speed it's going up at, at a certain height, is the same it is coming down at, at the SAME height.

    Hence figure how fast using you V2 = (Vo*Sinθ)2 - 2*g*y

    where y is your height.

    Once you have the 2 components of velocity let Pythagoras figure the rest for you.
     
  5. Jan 25, 2009 #4
    Thanks! Got it!
     
  6. Jan 25, 2009 #5

    LowlyPion

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    Great. Good luck.
     
  7. Feb 16, 2009 #6
    I need help getting started on this problem. The book doesn't really explain where to go so please help!

    Jan
     
  8. Feb 16, 2009 #7

    LowlyPion

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    Welcome to PF.

    The angle is not given, but they give you how to get it.

    I needs to go 45 m in 3 s. That means that the horizontal velocity = VoCosθ = 45/3 = 15. If VoCosθ = 15 then Cosθ = 15/25

    That gives you your angle.

    It also then gives you your vertical component of velocity, because that is VoSinθ.

    With vertical velocity you can figure the Vy at the window from

    Vy = Voy - g*3

    where 3 is the time to get there.

    With the vertical and horizontal components of velocity known you can determine |Velocity| at the window.
     
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