Derivation of Simple Projectile Motion with Drag Given the initial velocity $v_0$ and angle $\theta$ of a projectile on the ground, using Newton's second law and the acceleration due to gravity $\mathbf g=\left\langle0,-g\right\rangle$, I was able to derive its position vector function:
$$
\mathbf F=m\mathbf a=m\mathbf g\implies\mathbf r(t)=\left(v_0t\cos\theta,-\frac g2t^2+v_0t\sin\theta\right).
$$
I now want to introduce drag into this function. From my differential equations book, I was able to deduce that
$$
m\mathbf a+c\mathbf v=m\mathbf g,
$$
Where $c$ is some scalar. Is this correct? If so, how can I go about solving for $\mathbf r$? Integrating factors make no sense to me in this context.

I think (I can't actually remember at the moment) that the dragging factor is certainly proportional to the speed. So, the last equation you wrote yields two ordinary differential equations:
\begin{align*}
&\frac{d^2}{dt^2}r_x(t) + \frac{c}{m}\frac{d}{dt}r_x(t) = 0\\
&\frac{d^2}{dt^2}r_y(t) + \frac{c}{m}\frac{d}{dt}r_y(t) + g = 0
\end{align*}
To solve the first one, just observe that:
$$
\frac{d^2}{dt^2}r_x(t) + \frac{c}{m}\frac{d}{dt}r_x(t) = \frac{d}{dt}\left(\frac{d}{dt}r_x(t) + \frac{c}{m}r_x(t) \right)=0,
$$
which means that $\frac{d}{dt}r_x(t) + \frac{c}{m}r_x(t)$ is constant in time. Do you know how to solve this?
To solve the second equation, observe that it's the same as the first one but with the constant term added. Remember that the general solution of an inhomogeneous ODE, is the sum of the general solution and one particular solution (for example, $r_y(t) =-\frac{gm}{c}t$ is a particular solution).
Let me know if something isn't clear.
