With the drag force
$$
{\bf{F}}_d = - \alpha \left|{\dot{\bf r}}\right|^2 \frac{{\dot{\bf r}}}{\left|{\dot{\bf r}}\right|} = - \alpha \left|{\dot{\bf r}}\right| {\dot{\bf r}},
$$
where $\alpha = \rho C_d A / 2$, the equation of motion for the cannonball is
$$
m {\ddot{\bf r}} = - m g \ {\hat{\bf y}} + {\bf{F}}_d = - m g \ {\hat{\bf y}} - \alpha \left|{\dot{\bf r}}\right| {\dot{\bf r}},
$$
or
$$
\begin{align}
{\ddot{x}} &= f_x\left({\dot x}, {\dot y}\right) = - \beta {\dot{x}} \sqrt{{\dot x}^2+{\dot y}^2} \\
{\ddot{y}} &= f_y\left({\dot x}, {\dot y}\right) = - g - \beta {\dot{y}} \sqrt{{\dot x}^2+{\dot y}^2}
\end{align}
$$
where $\beta = \alpha / m = \rho C_d A / 2 m$.
Given initial conditions $x\left(0\right)$, $y\left(0\right)$, ${\dot x}\left(0\right)$, and ${\dot{y}}\left(0\right)$, you can then numerically integrate the system of equations ${\ddot{x}} = f_x\left({\dot x}, {\dot y}\right) $, ${\ddot{y}} = f_y\left({\dot x}, {\dot y}\right)$.