 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 givesThe downward terminal velocity is $-v_t$, whereEquivalently, if the drag acceleration is specified as $-k\mathbf{v}$, then $v_t=g/k$. Solving the first-order equation,The highest point occurs at $v_y=0$, soFor an upward launch this root is positive and ${\dot{v}}_y=-g$ there, confirming a maximum height. As $b\to 0$, expanding the logarithm gives the drag-free result $t_{\mathrm{top}}\to u_0\sin \alpha /g$. A horizontal launch has its highest point at launch; a downward launch from the ground does not have a later airborne maximum.