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Projectile motion(?) solving for Vo

  1. Sep 24, 2006 #1
    How do I find initial velocity if I'm only given time and the angle above the horizontal?
    The only other info given is that the object is caught at the same level it is kicked.
     
  2. jcsd
  3. Sep 24, 2006 #2

    radou

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    What time are you given? The time 'caught as the same level it was kicked'? You mean, the same vertical position?
     
  4. Sep 24, 2006 #3
    I think so. >.< My book gives me equations but they all ask for V sub y and Vo but I dont have either...
     
  5. Sep 24, 2006 #4

    radou

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    You have to use the equation for the displacement of the object in the y-direction, y(t0) = v0*sinA*t0-1/2*g*t0^2. Now, think about what y(t0) has to equal at the time t0, at which it has reached the point with the same y-coordinate as the one from which it was thrown. Then you can directly get the initial velocity v0 from the equation above, with a given angle A and time t0.
     
  6. Sep 24, 2006 #5
    I don't understand. I don't know v0 so how do I solve the equation?
     
  7. Sep 24, 2006 #6

    radou

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    Of course you can solve the equation and get v0 by setting v0*sinA*t0-1/2*g*t0^2 = 0.
     
  8. Sep 24, 2006 #7
    I don't really know how to rearrange equations, but would this be right? thanks
    Vo =t0-1/2*g*t0^2/sinA
     
  9. Sep 24, 2006 #8

    radou

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    [tex]v_{0}=\frac{\frac{1}{2}gt_{0}^2}{sin \alpha \cdot t_{0}}[/tex]. Was this so hard? :smile:
     
  10. Sep 24, 2006 #9
    :D Got it! Thanks for the help I really appreciate it. I'm a really slow learner lol. How did t0 end up at the bottom?
     
  11. Sep 24, 2006 #10

    radou

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    We divided by it. :smile:
     
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