$$ a_{xd} = k({v_x}^2+{v_y}^2)cos\beta $$ $$ a_{yd} = k({v_x}^2+{v_y}^2)sin\beta $$
According to the above equations (relating to a projectile experiencing drag), the drag acceleration components are dependent on both the parallel AND the perpendicular velocity components. I'm struggling to understand how, say, the horizontal drag acceleration could depend on the vertical velocity.
EDIT:
For reference, the equations have been derived for a projectile as follows: $$ a_d=\frac{1}{2m}C_d\rho A V^2 $$ where $C_d$ is the drag coefficient, $\rho$ is the medium density, A is the projected area of the body and V is the velocity.
For the purpose of these calculations: $$ k = \frac{1}{2m}C_d\rho A $$ Therefore the magnitude of the drag acceleration can be expressed as: $$ |a_d| = k ({v_x}^2+ {v_y}^2) $$ If the angle of elevation beta is expressed as: $$ \beta = \arctan{\frac{v_y}{v_x}} $$ Then the x and y components for the drag acceleration can be expressed as in the image above.