If we assume the drag force is quadratic, that is, $\vec{F_{d}} \propto v^2 \hat{v}$, then $F_x \propto (v_x^2 + v_y^2).$ My question is not why this happens mathematically, but rather, why does it make sense physically that the drag force in the $x$ direction should depend upon the speed in the $y$ direction? Why shouldn't the equation instead be: $$\vec{F_{d}} \propto v_x^2 \hat{x} + v_y^2 \hat{y}?$$


Physically, the drag force cannot depend on how you set up your coordinate system. That means that the force must transform under rotations like a vector, and it must be formed out of vector and scalar quantities. The only vector (besides $\vec{F}$ itself) in the drag problem is $\vec{v}$, and the scalars are $v$, as well as material properties of the air like density and viscosity.

For the particular case of a quadratic drag, this means that the only possibility is $\vec{F}\propto v^{2}\vec{v}$; there is no other vector in the problem that can specify the direction. What is wrong with the proposed alternative you give is that if you choose different coordinates, you get a different answer. If you set up new coordinate axes at a 45-degree angle to the $x$- and $y$-axes, you will not find a force with the same form you wrote down.

The mechanism for an aerodynamic drag force that depends on a power of the speed greater than 1 is that energy is lost in the process of forcing the fluid medium out of the way of the moving body. A projectile moving in the $x$-direction pushes fluid out of the way in the perpendicular $y$- and $z$-directions, as well as producing vortices of rotating fluid material. So the body's motion in one direction affects the motion of the medium along other directions, and hence affects the response of the medium to a body moving in a different direction.

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