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Parabolic Equation assistance please.

  1. May 19, 2009 #1
    I've been having difficulty understanding how the Parabolic Equation works... everytime I do a problem I get some really nasty looking numbers and often times it just doesn't work at all.
    The equation I was taught is:

    y = tan(θ)x - ( g / 2vo²cos²θ ) where vo = initial velocity vector

    This equation was derived from: x = VxoT -or- T = x / Vxo (T = time) and
    y = VyoT - ½gt²

    Please help me understand how to apply this formula correctly.

    Thank you very much,
    Greg from Mass.
    p.s. I know I didn't answer part three of the Template, however my question is more conceptual based rather than example. Pls don't delete
     
  2. jcsd
  3. May 19, 2009 #2

    rock.freak667

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    Homework Helper

    Well for a given angle and initial velocity, you can find the vertical position at any horizontal position. I'm not too really sure what kind of answer you are looking for.
     
  4. May 19, 2009 #3

    berkeman

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    Staff: Mentor

    Welcome to the PF, Greg. I'm not seeing where your top equation comes from. The bottom ones appear to be the kinematic equations of motion for constant acceleration (the "g" in this case), although they are a bit difficult to read without using LaTex:

    [tex]y(t) = y_0 - \frac{1}{2} gt^2[/tex]

    How did you go from the kinematic equations to your first equation?
     
  5. May 19, 2009 #4
    Thanks for the responses so quickly. And I"m sorry but I don't know what LaTex is, or where to obtain it. :(

    My textbook tends to be super confusing about stuff that should be pretty simple, I guess.
    The way it's described is:

    t = x/Vxo is substituted into y = VyoT - ½gt² to obtain
    y = Vyo(x/Vxo) - ½g(x/Vxo)² which can be rewritten as
    y = (Vyo/Vxo)x - (g/2v2ocos2Θo)x2 or
    y = (tanΘo)x - (g/2V2ocos2Θo)x2

    and V2o is supposed to be initial velocity V squared.
     
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