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.