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Ballistic problem: reachable region

  1. Jul 2, 2012 #1

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  3. Jul 2, 2012 #2

    tiny-tim

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    no, it looks like the 2D region of all points that the projectile can pass through with initial speed vo
     
  4. Jul 2, 2012 #3
    Thanks! could you please explain further how you know this to be true?
    If you look here: http://en.wikipedia.org/wiki/Trajectory#Range_and_height
    I'm assuming x is the max range R?
    Then y = vi2sin2(θ)/2g = vi2/2g * (1 - cos2(θ))
    Now if this were to match the equation in OP, then we need
    vi2cos2(θ)/2g = gR2/2vi2
    But subbing in R = vi2sin(2θ)/g, we get
    vi2cos2(θ)/2g = vi2sin2(2θ)/2g
    But cos2(θ) =/= sin2(2θ)
     
    Last edited: Jul 2, 2012
  5. Jul 3, 2012 #4

    tiny-tim

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    hi snellslaw! :smile:
    where does this come from? surely y = 0 ? :confused:

    if you put x = v2sin(2θ)/g into the equation, and θ = 45°, you do get y = 0 :wink:
     
  6. Jul 3, 2012 #5
    Thanks tiny-tim! :D
    I think the line you quoted was not my question however;
    we need vi2cos2(θ)/2g = gR2/2vi2
    but this leads to cos2(θ) = sin2(2θ) which is not an equality.

    Thanks again!!
     
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