In the presence of linear drag, the equation of motion is
\begin{align*}
\dot{\mathbf{v}} + 2\gamma \mathbf{v} = \mathbf{g} \implies
\begin{cases}
\dot{v}_x + 2\gamma v_x &= 0 \\
\dot{v}_y + 2\gamma v_y &= -g
\end{cases}
\end{align*}A solution to the homogenous equation is ##e^{-2\gamma t}##. The forced equation for ##v_y## has a particular solution of ##-g/2\gamma##. The velocity is\begin{align*}
\mathbf{v}(t) = \begin{pmatrix} ae^{-2\gamma t} \\ be^{-2\gamma t} - \frac{g}{2\gamma} \end{pmatrix}
\end{align*}If ##\mathbf{v}(0) = (v_{\mathrm{0x}}, v_{\mathrm{0y}})## then ##a = v_{\mathrm{0x}}## and also ##b = g/2\gamma + v_{\mathrm{0y}}##, so that\begin{align*}
\mathbf{v}(t) &= e^{-2\gamma t} \begin{pmatrix} v_{\mathrm{0x}} \\ v_{\mathrm{0y}} + \frac{g}{2\gamma}[1- e^{2\gamma t}] \end{pmatrix}
\end{align*}Integrating, and putting ##\mathbf{x}(0) = 0##, gives\begin{align*}
\mathbf{x}(t) &= \frac{1}{2\gamma} (1- e^{-2\gamma t}) \begin{pmatrix} v_{\mathrm{0x}} \\ v_{\mathrm{0y}} + \frac{g}{2\gamma} \left[ 1 - \frac{2\gamma t}{1- e^{-2\gamma t}} \right] \end{pmatrix}
\end{align*}