 Projectile motion with linear drag

For constant acceleration vector $\mathbf{g}$ and $\gamma >0$, a projectile initially at ${\mathbf{r}}_0$ with velocity ${\mathbf{v}}_0$ hasThe integrating factor $e^{\gamma t}$ solves the first-order linear ordinary differential equation for the velocity; one more integration proves the position formula. The terminal velocity is $\mathbf{g}/\gamma$, and $1/\gamma$ is the relaxation time. For $\mathbf{g}=(0,-g)$ and ${\mathbf{r}}_0=0$, eliminating time when $V_{0x}\ne 0$ givesFor finite $t\ge 0$, the physical domain is $0\le \gamma x/V_{0x}<1$. If $V_{0x}=0$, the path is vertical and cannot generally be represented as a single-valued graph $z(x)$. Expanding at fixed $x$ as $\gamma \to 0$ recovers $z=(V_{0z}/V_{0x})x-g{x}^2/(2V_{0x}^2)$.