 Solution

For a launch angle $\theta$, the projectile motion equations give a flight time $2u\sin \theta /g$, a projectile range $R=u^2\sin (2\theta )/g$, and an apex height $u^2{\sin }^2\theta /(2g)$. Thus the ceiling requires ${\sin }^2\theta \le 2gH/u^2$.

Set $z={\sin }^2\theta$. Then $R^2=4u^{4}z(1-z)/g^2$, which increases up to $z=1/2$. The optimum is therefore $z=\min (1/2,2gH/u^2)$, giving the projectile range under a ceilingThe printed expression is valid in the first regime. A higher ceiling leaves the usual $45^{\circ }$ launch admissible, so it cannot reduce the maximum range. The angle in the ceiling-limited regime is $\arcsin \sqrt{2gH/u^2}$. Equality at the roof is allowed in this ideal geometrical model; if touching the roof is forbidden, the same value is a supremum approached from below.