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Parabolic Path of a Projectile

  1. Jul 11, 2009 #1
    How do we know, or how can we prove that the path of a projectile is (at least part of) a parabola?
     
  2. jcsd
  3. Jul 11, 2009 #2

    nicksauce

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    Easily:

    In the vertical direction
    [tex]
    F = ma = m\frac{d^2 y}{dt^2} = -mg
    [/tex]

    Integrated twice we obtain
    [tex]
    y(t) = y(0) + v_y(0)t - \frac{1}{2}gt^2
    [/tex]

    Which is indeed the equation of a parabola with respect to time.

    Now in the x-direction, we have
    [tex]
    x(t) = x(0) + v_x(0)t
    [/tex]
    since there is no force in this direction. It should be clear that if we re-write y(t) in terms of x(t), by solving for t in the above equation, y(t) will also be a parabola in terms of x.
     
  4. Jul 11, 2009 #3

    rcgldr

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    A parabola occurs under the simplified case of a flat earth and a constant gravitational field and no aerodynamic drag. For a spherical earth, with no drag, and the equivalent of a point source for gravity, the path is part of an ellipse.
     
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