= Solution
Write the <linear drag> as $\mathbf F_d=-b\mathbf v$, with $b>0$, and let $m$ be the projectile's <mass>. Taking upward as positive, <Newton's second law> gives
$$
m\dot v_y=-mg-bv_y,\qquad v_y(0)=u_0\sin\alpha.
$$
The downward <terminal velocity> is $-v_t$, where
$$
\boxed{v_t=\frac{mg}{b}.}
$$
Equivalently, if the drag <acceleration> is specified as $-k\mathbf v$, then $v_t=g/k$. Solving the first-order equation,
$$
v_y(t)=(u_0\sin\alpha+v_t)e^{-bt/m}-v_t.
$$
The highest point occurs at $v_y=0$, so
$$
\boxed{t_{\rm top}=\frac mb\ln\left(1+\frac{u_0\sin\alpha}{v_t}\right)
=\frac{v_t}{g}\ln\left(1+\frac{u_0\sin\alpha}{v_t}\right).}
$$
For an upward launch this root is positive and $\dot v_y=-g$ there, confirming a maximum height. As $b\to0$, expanding the logarithm gives the drag-free result $t_{\rm top}\to u_0\sin\alpha/g$. A horizontal launch has its highest point at launch; a downward launch from the ground does not have a later airborne maximum.
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