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Phsyics Kinematics: cannonball shot straight up

  1. Feb 28, 2010 #1
  2. jcsd
  3. Feb 28, 2010 #2
    here is my math for it.

    [tex]v_{y}=v\sin\theta[/tex]
    [tex]v_{x}=v\cos\theta[/tex]

    [tex]t=\frac{v\sin\theta}{g}[/tex]
    to get the total time in the air multiply by two, becuase that time is only half way.
    [tex]x=v\cos\left(\theta\right) t[/tex]
    plug in t
    [tex]x=\frac{2v^2\sin\theta \cos\theta}{g}[/tex]

    that should get you the distance traveled... i think but look over my math.
     
  4. Mar 1, 2010 #3
    Oh. I'm having trouble with the angle though. Not sure what it would be.
     
  5. Mar 1, 2010 #4
    and isn't there acceleartion in the x direction too?
     
  6. Mar 1, 2010 #5
    From what i saw in the problem there were not any more forces than just gravity... and gravity only works in the y direction. your angle is the one given in the problem, 45.
     
  7. Mar 1, 2010 #6
    Oh I was talking about question 23. Hehe I already solved for question 22. Oh well. My exam is in 20 min. Dumdumdum. So it's okay. Thanks for the help though
     
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