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Under uniform gravity without drag, a projectile launched at speed $u$ from height $h$ reaches a maximum horizontal projectile range $u\sqrt{u^2+2gh}/g$. For horizontal position $X$, maximize $y=h+Xz-g{X}^2(1+z^2)/(2u^2)$ over $z=\tan \alpha$. The maximizer is $z=u^2/(gX)$ and its envelope meets the ground at the displayed distance. The optimum angle obeys $\tan \alpha =u/\sqrt{u^2+2gh}$.

For maximum projectile range from an elevated launch, let $X$ be horizontal projectile range from the tower and $\alpha$ the launch angle above horizontal. The projectile motion trajectory isPut $z=\tan \alpha$. For a fixed $X$, its largest possible $y$ occurs at $z=u^2/(gX)$; the envelope is $y_{\max }=h+u^2/(2g)-g{X}^2/(2u^2)$. The furthest attainable ground point satisfies $y_{\max }=0$, soThe optimum angle is less than $45^{\circ }$ for $h>0$ and approaches $45^{\circ }$ as $h\to 0$. The result is the maximum horizontal distance, using uniform gravity and no drag.

Use coordinates $s$ down the inclined plane and $n$ along its upward normal, with the launch point as origin. Gravity has components $g\sin \varphi$ and $-g\cos \varphi$. For launch angle $\theta$ above the downhill direction,The nonzero root of $n(t)=0$ is $t_f=2v_0\sin \theta /(g\cos \varphi )$. Substituting into $s(t)$ and combining the two trigonometric terms gives the distance along the slope:This assumes $0\le \varphi <\pi /2$, $\sin \theta >0$, and a downhill landing with $\cos (\theta -\varphi )>0$, as well as the neglected air resistance. It reduces to the horizontal projectile range $v_0^2\sin (2\theta )/g$ at $\varphi =0$. The angle relative to the horizontal is $\theta -\varphi$, which explains the shifted cosine. The original diagram uses stipulated illustrative angles; no examination figure is being reconstructed.

Take the launch point as $(x,y)=(0,0)$ and ignore air drag. In projectile motion with constant gravitational acceleration $g$, a launch at angle $\theta$ above the horizontal givesAt the wall $x=l$, the height isThis is a concave quadratic polynomial in $q$, so its unique maximum satisfies $l-g{l}^{2}q/v_0^2=0$. HenceFor $l=v_0^2/g$, the best trajectory has $\theta =45^{\circ }$ and reaches the wall at launch height. For $l>v_0^2/g$, the mathematical maximum height is negative. If the ball must stay above level ground, it lands before reaching the wall: $v_0^2/g$ is the maximum same-level projectile range. If there is no intervening ground and the wall extends below the launch height, the same optimum still applies, but its impact point is below that height.

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.

A projectile launched from and returning to the floor beneath a horizontal ceiling of height $H$ has maximum projectile range $\sqrt{8H(u^2/g-2H)}$ when $0\le H\le u^2/(4g)$. For higher ceilings the maximum is $u^2/g$. This assumes uniform gravitational acceleration, no air resistance and fixed launch speed $u$.

Indeed, with $z={\sin }^2\theta$, the apex condition is $z\le 2gH/u^2$ and $R^2=4u^{4}z(1-z)/g^2$. This concave quadratic function is maximized at the admissible value nearest $1/2$. Thus the lower-ceiling expression must not be extrapolated past its regime. If touching the ceiling is excluded, the limiting value is a supremum instead of an attained maximum.