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Ballistic Range FOrmula

  1. Feb 19, 2008 #1
    I completely understand the mathematics behind finding the range of a projectile flying over flat ground. But I cannot find a reliable formula for the projectiles range if it is flying over non-flat ground. If someone knows the correct formula and could perhaps explain why it works, it would be most appreciated!
  2. jcsd
  3. Feb 20, 2008 #2
    To "completely understand..." would imply that you know how to construct the trajectory of the projectile.

    Simple -- just compute the first intersection of the projectile trajectory and ground surface :)

    Having said that, it is clear there can be no "correct formula" in general case, because the ground surface may be any kind of function, such that there is no analytical way to compute the intersection.

    Chusslove Illich (Часлав Илић)
  4. Feb 20, 2008 #3


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    Assuming air resistance is ignored, and you have constant acceleration...

    [tex]x(t) = v\cos(\theta)t [/tex]

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

    which gives the range (after some algebraic manipulation),

    [tex] d = \frac{v\cos(\theta)}{g} \cdot [v\sin(\theta) + \sqrt{[v\sin(\theta)]^2 + 2gy_0}][/tex]


    [tex] \theta [/tex] is the launch angle
    [tex] y_0 [/tex] is the initial launch height
    v is the launch velocity
    d is the horizontal distance the projectile will travel (i.e. the range)

  5. Feb 20, 2008 #4
    thanks guys, that helps a lot!
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