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It's often stated that motion in the $x$ and $y$ axises are independent, so that changing the $x$-velocity will not influence changes in the $y$-velocity. To me it seems that with quadratic drag (drag proportional to $v^2$) this shouldn't hold true. If we increase the initial velocity in the $y$-direction, this would increase the drag. It would also decrease the percent of the drag being applied in the $x$-direction, since the drag acts opposite to the velocity. Do these effects cancel?

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    $\begingroup$ 2D quadratic drag was also considered in this Phys.SE post and links therein. $\endgroup$ – Qmechanic Jun 30 '15 at 23:53
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So the x part of the drag is

$$\begin{align*} f_x &= f \cos\theta \\ &= (k v^2) \cos\theta \\ &= k v (v \cos\theta) \\ &= k v v_x \,, \end{align*}$$

and $v$ is dependent on the $y$ component of motion as well as the x-component.

Similar consideration, of course, apply to the y-component of drag.

So the independence is explicitly lost and you can no longer solve each direction without reference to the other. That was a helpful feature of the introductory course not present in the full problem.


Note that in a regime where $f \propto v$ you can retain the independence of the two coordinate equations. I conjecture that this is why that fairly rare case is often treated in upper-division mechanics texts.

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  • $\begingroup$ Thank you very much, you've made me a happy betting man! Would this mean that increasing the $x$ component actually would decrease the maximum height reached, since it would increase the amount of drag? $\endgroup$ – vxstraza Jul 1 '15 at 1:41
  • $\begingroup$ You should look at the link in Qmechanic's comment. The details of this problem have been of interest for a long time and progress on non-numerical solutions has been made in recent decades. Mind you, the numerical solutions have been good enough for gunnery for more than a century. $\endgroup$ – dmckee Jul 1 '15 at 1:44

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