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Horizontal projection

  1. Sep 20, 2008 #1
    1. The problem statement, all variables and given/known data

    A firefighting crew uses a water cannon that shoots water at 25.0 m/s at a fixed angle of 53.0 degrees above the horizontal. The firefighters want to direct the water at a blaze that is 10.0 m above ground level.

    How far from the building should they position their cannon? There are two possibilities (d_1<d_2); can you get them both? (Hint: Start with a sketch showing the trajectory of the water.)

    2. Relevant equations

    x=vix*t
    vfx=vix
    vfx^2=vix^2
    y=viy*t+0.5a8t^2
    Vfy=viy+a*t
    vfy^2=viy^2+2*a*y
    3. The attempt at a solution

    i do not know where to start with this problem. Please and Thank you for the help!!
     
  2. jcsd
  3. Sep 20, 2008 #2

    LowlyPion

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    Homework Helper

    What is the equation for the motion of the water through the air in the y direction and the x direction?

    At what times will the height of the water satisfy the y position equation?

    What does that mean in terms of how far away you need to be using your x direction equation?
     
  4. Sep 21, 2008 #3
    Knowns:
    y component

    Yo=0
    y=10
    vyo=25sin53
    =19.97
    a=9.8

    X component

    xo=?
    xf=0
    vxo=25sin53
    t=
    a=0


    y=yo+voyt-0.5at^2 10=25sin53t-0.5 9.8 t^2 it can form the quadratic equation so -4.9t^2+25sin53t-10 t=0.58 and 3.49

    and i take those numbers and enter them into 25cos53 t =x i get 8.7 and 51.3 but the answer that has 3.49 seconds is wrong. Does anybody know what i am doing wrong?
     
  5. Sep 21, 2008 #4

    LowlyPion

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    Homework Helper

    Using .6018 as Cos 53, I get 15.04 as horizontal velocity and the 3.49 yields 52.5 m. And 8.8 for the closer one.
     
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