= Solution
First solve for the <velocity> components. The <integrating factor> $e^{\gamma t}$ gives
$$
\dot x=V_{0x}e^{-\gamma t},\qquad
\dot z=\left(V_{0z}+\frac g\gamma\right)e^{-\gamma t}-\frac g\gamma.
$$
Integrating once more and imposing the initial position gives
$$
\boxed{x(t)=\frac{V_{0x}}\gamma(1-e^{-\gamma t}),\qquad
z(t)=\frac{V_{0z}+g/\gamma}\gamma(1-e^{-\gamma t})-\frac g\gamma t}.
$$
These formulas describe <projectile motion with linear drag>. Differentiation checks both second-order <differential equations> and both initial velocities. At large positive <time>, the horizontal <velocity> tends to zero and the vertical <velocity> tends to the terminal value $-g/\gamma$.
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