= Solution
For positive <linear drag>, $v_x$ decays exponentially toward zero. The vertical <velocity> decreases monotonically toward $-mg/\alpha$, with an initially steeper gradient $-g-(\alpha/m)v_0\sin\theta$ and positive second <time derivative> for an upward launch. It crosses zero at
$$
\boxed{t_{\rm top}=\frac m\alpha\ln\left(1+\frac{\alpha v_0\sin\theta}{mg}\right).}
$$
The right panel shows the two components and the formal <terminal velocity>. To identify the end of the physical flight, integrate the vertical <velocity> once:
$$
y(t)=\tau_d(v_0\sin\theta+g\tau_d)(1-e^{-t/\tau_d})-g\tau_dt.
$$
The first positive root of $y(t)=0$ is ground contact. The <terminal velocity> need not be reached before that event; dashed portions in the sketch are mathematical extensions without a ground surface.
\Image[/past-exam-of-the-physics-1-paper-of-the-university-of-oxford/2017-projectile-velocity.png]
{title=Projectile velocity components with and without linear drag}
{description=An illustrative launch at $45^\circ$, with dimensionless time $T=gt/v_0$. The right panel uses $\alpha v_0/(mg)=1$. Solid curves end at the first return to the launch height; dashed curves show the free-flight continuation.}
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