Understanding projectile motion formula with air resistance

Question

$\frac{dx}{dt} = v_x$

$\frac{dy}{dt} = v_y$

$\frac{dv_x}{dt} = -b|v|v_x$

$\frac{dv_y}{dt} = -g -b|v|v_y$

where $v = \sqrt{v^2_x+v^2_y}$ and $b$ is a drag constant

Why is it that the magnitude of the speed (used in the last two equations) is multiplied with the speed in the $y$ and $x$ direction?

Answer 1

According to your information, air resistance $\mathbf{f} \propto -v^{2}$ and \begin{align*} \frac{\mathbf{f}}{m} &= -bv^{2} \hat{\mathbf{v}} \\ &= -bv^{2} \frac{v_{x}\, \mathbf{i}+v_{y}\, \mathbf{j}} {|v|} \\ &= -b|v|(v_{x}\, \mathbf{i}+v_{y}\, \mathbf{j}) \\ \mathbf{a} &= \mathbf{g}+\frac{\mathbf{f}}{m} \\ &= -b|v|v_{x}\, \mathbf{i}-(g+b|v|v_{y})\, \mathbf{j} \end{align*}