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Homework Help: Projectle problem

  1. Sep 26, 2007 #1
    1. The problem statement, all variables and given/known data

    A cannon with a muzzle speed of 992 m/s is used to start an avalanche on a mountain slope. The target is 1900 m from the cannon horizontally and 793 m above the cannon. At what angle, above the horizontal, should the cannon be fired? (Ignore air resistance.)

    2. Relevant equations

    Xf= xi+vit
    Yf=yi+vit+1/2at2
    vfy^2=viy^2+2a(yf-yi)
    x=cos(theta)992m/s
    Y=sin(theta)992m/s

    3. The attempt at a solution
    ok this one i can get all the equations setup.
    xf= cos(theta)992m/s
    793=(sin(theta)992m/s)t + 1/2(-9.80)t2
    using the x eqaution i can solve for t.
    t=1900/cos(theata)992m/s
    but once i place it into the eqaution, it gets very messy and i get stuck on the arithmatic.
    793= (sin(theta)992m/s)(1900/cos(theata)992m/s)+1/2(-9.80)(1900/cos(theata)992)^2
    this is where i am stuck.
     
  2. jcsd
  3. Sep 26, 2007 #2

    Avodyne

    User Avatar
    Science Advisor

    You have a position measured in meters per second. This obviously doesn't make sense.
     
  4. Sep 26, 2007 #3
    i know the it shuold look like this
    xf=xi+vit
    1900m= 0+(cos(theta)(992m/s)t
     
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