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Max Height and Time

  1. Jun 7, 2012 #1
    I am a formal instructor and one of the classes I teach revolves
    around the ballistics of a bullet. I am currently in a Calculus class
    and will be starting as an Economics major In the fall when I finish
    my time in the Marines. I am hoping to see if there is any way,
    through calculus that I can prove the max height of the round through
    actual mathematical proof. Many of the lesson plans and outlines
    simply indentify what I have attached below and do not mention proof
    of this. I am simply curious if I can prove the max height of the
    round and at what time during its flight path does it reach this
    height. Any help would be greatly appreciated. Below is a simple
    diagram of a rifle trajectory at 300 yards. I attached the image to
    give a preliminary view of what I am referencing.

    Maximum Ordinate - The highest point in the trajectory of the round
    on its route to the target
    The event I would like to know if I can prove is how high the maximum
    ordinate should be. I am new to calculus and especially the
    application of calculus to real world events. Can I use a Derivative
    function formula to find the max?

    If so, what variables do I need for this, the rounds are all constant,
    and with that the velocity of the round at specific ranges is also
    fixed (in theory).
    Would I use a function like h(t) = 4t^2+48t+3 to find the maximum h
    and the corresponding time t. The conventional knowledge says that
    roughly 2/3rds of the way to the target the projectile reaches its
    maximum height. It is said to be roughly 7 inches above the line of
    sight at its highest moment, which is supposedly 2/3rd of the way to 300 yards. This is what I am trying to prove or disprove throughcalculus.
    All in all, any help or nudges to this type of problem would be
    greatly appreciated.
  2. jcsd
  3. Jun 7, 2012 #2
    This seems like an appropriate problem for calculus, but I'm unsure of what assumptions are necessary to get the correct result. In other words, it's usually an introductory physics problem to treat the trajectory of an object under the influence of uniform gravity without drag; in that scenario, the maximum height always occurs at halfway through the trajectory. I suspect drag is needed for this problem, however.

    In other words, you'll have to solve a vector differential equation:

    [tex]\frac{d\vec p}{dt} = \vec F_{\text{drag}} + \vec F_{\text{gravity}}[/tex]

    The gravitational force is just going to be downward, but the drag will generally be in the opposite direction of the velocity. This is probably a tough problem to attack analytically, to be honest, which is why (I suspect) it's not often explicitly proven.
  4. Jun 7, 2012 #3
    Thanks for your response. I am going to have to do some research and reading to figure out a way to solve this. I have come across many issues with some of the material I teach that should be able to be proven mathematically.

    You mention that max height always occurs halfway through trajectory, but in this case all the literature teaches that 2/3rds of the way to the target a bullet will reach its max ordinate, or highest point.
  5. Jun 8, 2012 #4


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    Science Advisor

    No, he said that without drag "the maximum height always occurs at halfway through the trajectory". With drag, which reduces the horizontal speed of the bullet more than the vertical speed, the trajectory is steeper later than before and so the maximum height is farther on the trajectory. What formula for "drag" is assumed in the literature that gives 2/3? Both "proportional to the speed" and "proportional to the square of the speed" are often used and will give different results.
  6. Jun 10, 2012 #5
    I appreciate the input. As i might have not mentioned, I am new to the world of calculus, and especially new to the world of attempting to apply it to a real life scenario. I teach marksmanship, and the pubs are simply what our organization gives us for lesson plans. The lesson plan says 2/3rds of the way to the target at 300 yards, the round reaches roughly 7 inches above the line of sight.

    After the maximum ordinate (height) is taught and preached to be what I mentioned above, the lesson plan does indeed talk about drag and gravity.

    How would you guys, as the expert apply write a formula to prove or disprove when exactly over the 300 yards from end of barrel of rifle to impact of target that the round is at its highest over the line of sight? I have a table of the velocity of the round at specific distances. This is strictly for enjoyment of math and again I apologize if much of this is elementary, but I came to this sight to get input from those who have applied and mastered math in this form.

    Can anyone create a formula or throw a bone.

    Range is 300 yards

    the max ordinate is supposed to be at 200 yards (im assuming that is because of drag and external forces)

    The round weighs 62 gr and has velocity of 3,100 ft/s and is being shot out of a rifle with a muzzle velocity of 2900 ft per second.

    Am i missing any other important data? Do you need the ballistic coefficient of the round?
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