Though the mathematical concepts underlying quadratic air drag are quite straightforward (a single variable differential, just like the linear drag equation), my text book (and online text books) completely ignore the reasoning behind why the "drag" at high speeds on a trajectory through a fluid is proportional to the velocity squared. I know that this could not just be an empirical result.

Is there an intuitive reason for this (perhaps like momentum transfer or kinetic energy?). If possible, I would also like to ask for an intuitive reason behind the units $m^2/s^2.$

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## marked as duplicate by Ben Crowell, Brandon Enright, ACuriousMind, Ali, Qmechanic♦Aug 3 '14 at 0:53

What do you mean by "the units $m^2/s^2$"? –  Floris Aug 2 '14 at 21:35
Thanks Ben. Everyone had something very important to add for my understanding. It is much appreciated. –  Gödel Aug 2 '14 at 22:34

When you move fast enough, you create a "turbulent wake". It is as though the column of air that you push out of the way ends up traveling at your speed. This requires energy proportional to $v^2$ since the kinetic energy of that column will be proportional to $\frac12\rho v^2$. Thus the work done needed per unit distance is proportional to this quantity, and you need a force proportional to this quantity. Thus $F\varpropto v^2$ over a wide range of Reynolds numbers.
A body moving in a fluid will pass through fluid at a rate of $Av$ where $A$ is the cross-sectional area and $v$ the velocity. This clearly has units of volume per time. If the fluid is brought to rest relative to the body, the momentum transferred per unit time is $A\rho v^2$. But momentum transfer per unit time is force, so we expect that the drag force is proportional to $A\rho v^2$.