Resolve Drag into Vector Components Assume you have a sphere travelling at a velocity of $30$m/s, at angle $30^\circ$ relative to the local horizontal plane. For the purposes of the below equations please ignore more complicated influences on drag like turbulence.
It has a drag coefficient of $0.9$, and a frontal surface area of $0.2$m$^2$. I'm using a sphere because I assume the drag coefficient and frontal area is constant no matter the direction it travels. Therefore, using a basic formula for drag, this can be expressed by
$$D=\frac{1}{2} \cdot \rho \cdot A \cdot C_d \cdot V^2$$
My question is can you resolve this Drag force into its components by
$$D_x = D \cdot \cos\left(\frac{\pi}{6}\right)$$
$$D_y = D \cdot \sin\left(\frac{\pi}{6}\right)$$
Would this adequately resolve the drag force into its two components or do you need to resolve the velocity into its two components? e.g.
$$v_x = V \cdot \sin\left(\frac{\pi}{6}\right)$$
$$v_y = V \cdot \cos\left(\frac{\pi}{6}\right)$$
Then use those velocity components to calculate the drag acting in those directions? i.e.
$$D_x = \frac{1}{2} \cdot \rho \cdot A \cdot C_d \cdot (v_x)^2$$
$$D_y = \frac{1}{2} \cdot \rho \cdot A \cdot C_d \cdot (v_y)^2$$
I also assume the drag would be acting opposite the direction of the sphere, therefore become negative.
Are all of these interpretations correct? If they are incorrect please indicate which ones. 
 A: Drag acts in the opposite direction of velocity. Therefore, if
$$v_x = V \cdot \cos\left(\frac{\pi}{6}\right)$$
$$v_y= V \cdot \sin\left(\frac{\pi}{6}\right)$$
Then 
$$D_x = -D \cdot \cos\left(\frac{\pi}{6}\right)$$
$$D_y =-D \cdot \sin\left(\frac{\pi}{6}\right)$$
This is because, in general, if for vectors $\mathbf a$ and $\mathbf b$ the following is true
$$\mathbf b\propto-\mathbf a$$
then it's also true that
$$b_x\propto-a_x$$
$$b_y\propto-a_y$$
However, it is not true in your case that
$$D_x = \frac{1}{2} \cdot \rho \cdot A \cdot C_d \cdot (v_x)^2$$
$$D_y = \frac{1}{2} \cdot \rho \cdot A \cdot C_d \cdot (v_y)^2$$
You just need to take the drag force and break it into components using the correct trigonometric functions as done above. You don't replace all instances of $a$ with $a_i$ for each general vector $\mathbf a$.
A: In the general case, we can write the drag force in vector form
$$\vec F_D=\frac {1}{2}\rho A C_D(Re)|\vec v_0-\vec v|(\vec v_0-\vec v)$$
here $\vec v_0$ is wind speed, $\vec v$ is body speed, $Re$ is Reynolds number.
In the particular case of $\vec v_0=0$, we have
$$\vec F_D=-\frac {1}{2}\rho A C_D(Re)|\vec v|\vec v$$
The projections of the force on the coordinate axis have the form
$$(F_D)_{x}=-\frac {1}{2}\rho A C_D(Re)|\vec v|v_x, (F_D)_{y}=-\frac {1}{2}\rho A C_D(Re)|\vec v|v_y$$
In this case $|\vec v|=\sqrt {v_x^2+v_y^2}$. For a spherical particle, I can recommend the empirical formula for the drag coefficient
$$C_D=\frac {21.1}{Re}+\frac {6.3}{Re^{0.5}}+0.25$$
For $Re>>1$, we have $C_D=0.25$ (this is not 0.9).
