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Deriving a formula that an old physicist showed me.

  1. Sep 12, 2011 #1
    1. The problem statement, all variables and given/known data
    I was shown a formula by a physicist I met through a friend when I asked for help with a homework problem. I needed to find y final for a motion in 2 dimensions problem. He showed me this:

    Y1 = X1 * tan(Theta) + (Ay * X1^2) /( 2 * V0^2 * cos^2(Theta))

    Y1 = Y final
    X1 = X Final
    Ay = Acceleration of Y
    V0 = Initial Velocity

    I would love to know where this comes from and how to derive it. The X1tan(theta) makes sense to me: that would just = y of a triangle with base of x1. But the cos^2 and X1 ^2 confuses me. Did those come from an identity?

    Have any of you seen this before?
     
  2. jcsd
  3. Sep 12, 2011 #2

    rock.freak667

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    Homework Helper

    Take the simple components

    uyvsinθ and uxvcosθ (vertical & horizontal)

    When the projectile hits the ground y=0

    or y = uyt-1/2gt2 (u=initial and g = acceleration or your Ay)

    you can solve for t here.


    Then you know the entire horizontal range x=uxt, you know ux and t from above, so simplify.

    If you're having trouble, post your work.
     
  4. Sep 13, 2011 #3
    So I was able to derive the second half of the problem from 1/2at^2. The part that I am actually stuck on is the first chunk. I don't understand how ViT derives to XfTan(Theta)

    I got T = Xf/VCos(Theta)

    But somehow Viy*T = Xf Tan(Theta). Not sure how to derive that.
     
    Last edited: Sep 13, 2011
  5. Sep 13, 2011 #4

    rock.freak667

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    Homework Helper

    Now that I think it through, you are relating y with x.

    we have y=uyt-½gt2

    and we know that x=uxt or x=(vcosθ)t so t = x/vcosθ

    and now y = (vsinθ)t-½gt2

    sub t into it and you should get it.
     
  6. Sep 13, 2011 #5
    Thank you very much for your help. I figured it out. Couldn't have done it without help. :smile:
     
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