= Solution
Take $x$ horizontal, $y$ upward and $g>0$. Resolve <Newton's second law> into components for <projectile motion with linear drag>:
$$
m\dot v_x=-\alpha v_x,\qquad
m\dot v_y=-mg-\alpha v_y,
\qquad v_x(0)=v_0\cos\theta,\quad v_y(0)=v_0\sin\theta.
$$
For $\alpha>0$, let $\tau_d=m/\alpha$. An <integrating factor> $e^{t/\tau_d}$ gives
$$
\boxed{v_x(t)=v_0\cos\theta\,e^{-t/\tau_d},\qquad
v_y(t)=\left(v_0\sin\theta+g\tau_d\right)e^{-t/\tau_d}-g\tau_d.}
$$
The relaxation <time> is $\tau_d$, and the <terminal velocity> is $(0,-g\tau_d)$. For fixed finite $t$, expanding the <exponential function> as $\alpha\to0$ gives the continuous limit $v_x=v_0\cos\theta$ and $v_y=v_0\sin\theta-gt$; the two terms proportional to $\tau_d$ cancel before taking the limit.
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