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Trajectory of a bullet with drag

  1. Dec 3, 2008 #1
    Trajectory of a bullet with drag

    In most physics books, they pass over drag problems because of their difficulty. However, I had to do one for a mid-term for a physics class. When the velocity is low (tens of m/s or less), then the drag is proportional to the velocity and the math is quite simple (at least in comparison). However, we were asked about a bullet which travels nearly 1000 m/s and thus is dependent upon velocity squared. I was utterly confused with how to progress with the math.

    Here's what I had:

    Since the bullet is fired at an initially positive angle [tex]\theta[/tex], the drag can be broken into two components (2-D only): Fx and Fy.

    Fx = F * cos[tex]\theta[/tex]
    Fy = F * sin[tex]\theta[/tex]

    FDrag = [tex]\frac{(Drag Coefficient)*(\rho)*V^{2}*(Area)}{2}[/tex]
    FDrag = k*V2, where K = .5*Cd*[tex]\rho[/tex]*Area

    Assume that density is constant throughout time.

    Fx = m[tex]\frac{\partial vx}{\partial t}[/tex] = -k*V2

    [tex]\frac{\partial vx}{\partial t}[/tex] = -k*Vx2/m

    [tex]\frac{\partial vx}{V^2}[/tex] =-k/m * dt

    Integrate to get:

    -1/Vx = -kt/m + C

    Vx = m/kt + K

    However, at t = 0, Vx = muzzle velocity*cos(theta). But according to the equation I have, Vx would go to infinity as t -> 0. Right?

    I run into the same problem when trying to figure out Vy which is even more complicated since it involves gravity. Can anyone tell me where I went wrong with this?

  2. jcsd
  3. Dec 3, 2008 #2
    Re: Trajectory of a bullet with drag

    We'll get to your question, but I couldn't help to notice a bit of "creative math" in these two lines.

    [tex]\vec v^2 \neq v_x^2 [/tex]

    So in order to get the vec's right:

    [tex] m\frac{\partial \vec v}{\partial t}=-k \vec v^2 \hat v[/tex]

    (^ means unit vector) projecting onto the x-axis:

    [tex] m\frac{\partial v_x}{\partial t}=-k (v_x^2+v_y^2) \frac{v_x}{\sqrt{v_x^2+v_y^2}}[/tex]


    [tex] m\frac{\partial v_x}{\partial t}=-k v_x\sqrt{v_x^2+v_y^2}[/tex]

    and similar for y:

    [tex] m\frac{\partial v_y}{\partial t}=-k v_y\sqrt{v_x^2+v_y^2}[/tex]

    To get to your question, set vy = 0 for a case of one dimensional drag motion. Here:

    [tex] m\frac{\partial v_x}{\partial t}=-k v_x^2[/tex]

    Now you integrate to obtain
    Which is not a good (even wrong?) way to do it. Rather stick to:


    Where is it clear that v goes to v0 for t going to 0 for a suitable pick of the constant C (= 1/v0) and goes to 0 for t going to infinity.

    Now you can tackle the 2-dimensional problem on your own :)
  4. Dec 3, 2008 #3
    Re: Trajectory of a bullet with drag

    Wow, thanks. Not that complicated when you do it correctly.
  5. Dec 3, 2008 #4
    Re: Trajectory of a bullet with drag

    Check out these videos if you want your mind blow about drag:


    They might make you rethink the approximation that drag is proportional to the first or second power of the velocity.
    Last edited by a moderator: Apr 24, 2017
  6. Dec 3, 2008 #5
    Re: Trajectory of a bullet with drag

    Why? The gentleman in the videos drew nice parabolic drag/speed dependencies.

    Even if it is an approximation (it isn't, at least in some specific cases) who says that it is an unsatisfactory approximation?
    Last edited by a moderator: Apr 24, 2017
  7. Dec 3, 2008 #6
    Re: Trajectory of a bullet with drag

    Well the assumption does have some validity, for sure. But I think it would be naive to think that drag can be completely understood in these terms.
  8. Dec 3, 2008 #7


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  9. Dec 3, 2008 #8
    Re: Trajectory of a bullet with drag

    It's nice to find some physics where the theory is less developed than the experiment!
  10. Dec 4, 2008 #9
    Re: Trajectory of a bullet with drag

    First it was an approximation, now it is an assumption?? Please understand that it is neither. Just about any textbook on fluid mechanics will derive that. Of course there are limitations to the validity of the exact dependence, eg. in transition from sub-sonic to super-sonic. Such special cases will have to be tackled separately, just like is the case in just about all physics. But nobody has ever claimed otherwise.

    No. The fundamental theory of fluid mechanics (Euler (Navier-Stokes) equation + no slip BC) has proven to be remarkably robust, when solved correctly!

    It's time to call enough of this nonsense. If tankFan86 wishes to discuss disagreements in theory and experiment with regards to fluid drag, may I suggest that he provides a peer reviewed publication on which to base this discussion, such that we can all agree on the premises. This "I think..."-reasoning is useless.
    Last edited: Dec 4, 2008
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