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Newb Question

  1. Sep 4, 2007 #1
    Ball is kicked from location <6,0,-9> with initial velocity <-10,17,-6>. The ball's speed is low enough so that air resistance in negligible.

    What is the velocity of the ball after .5 seconds of being kicked?

    Use Momentum Principle


    I know that the x and z components will stay the same, but I do not understand how we can find the velocity if they do not give the mass. I keep coming up with

    [tex]_{V}iy[/tex]+(1/2)*([tex]_{F}y[/tex]/m)*([tex]\Delta[/tex]t)

    What am I missing?
     
  2. jcsd
  3. Sep 4, 2007 #2
    Here, the only force in the y direction is gravity, and f/m=g.
     
  4. Sep 4, 2007 #3
    Alright so f/m=g.

    V(yfinal)=[tex]_{V}iy[/tex]+(1/2)*(g)*([tex]\Delta[/tex]t)
    =<17>+[(1/2)(-9.80)]*.5
    =<17>+(-2.45)
    =14.55 m/s

    This answer is still incorrect. whats going on?
     
    Last edited: Sep 4, 2007
  5. Sep 4, 2007 #4
    I think F is acting in the z direction.
    and
    v(x,y,z) is = to <-10,17,..>+t<0,0,-g>
     
  6. Sep 4, 2007 #5
    I dont think so, my values for the x and z componenets are correct. Like chaos said, gravity is the only force acting on the ball, so that would be in the y direction.
     
  7. Sep 4, 2007 #6
    lol, you are using wrong formula

    vf = v0 + at

    because time is also 1/2, so i din't catch it. sorry :shy:
     
  8. Sep 5, 2007 #7
    aha! you got it rootx. good job thanks much.
     
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