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Use the arch geometry of two hinged rods sliding on a floor, with $A=O=(0,0)$ and $C$ on the other side of $B$. As long as $C$ remains in contact with the floor,The centres of mass of the two rods are $(l\cos \theta ,l\sin \theta )$ and $(3l\cos \theta ,l\sin \theta )$. Their squared speeds are respectively $l^2{\dot{\theta }}^2$ and $l^2(9{\sin }^2\theta +{\cos }^2\theta ){\dot{\theta }}^2$. Each uniform rod has moment of inertia $M{l}^2/3$ about its midpoint, and their angular velocities are $\dot{\theta }$ and $-\dot{\theta }$. Applying the kinetic energy decomposition about the center of mass to each rod givesThe gravitational potential energy is $U=2Mgl\sin \theta$. The fixed pivot does no work; neither does the frictionless floor reaction while $C$ slides horizontally. Conservation of energy from rest therefore givesOn the falling branch $\dot{\theta }$ is the negative square root. Just before the hinge reaches the floor, $\theta =0$, and its speed is $|\dot{B}|=2l|\dot{\theta }|$. Thus, under the maintained-contact interpretation,

There is a contact hypothesis behind this calculation. A floor can push upward but cannot pull downward. For $s=\sin \theta$ and $a=\sin \alpha$, the required normal reaction at $C$, found from the second rod's linear and angular equations, isFor the usual acute arch, this stays nonnegative throughout $0\le \theta \le \alpha$ exactly when $\sin \alpha \le 35/54$. Indeed, for $s>0$, its nonnegativity is $a\le 1/(18s)+11s/6+2s^3$, whose minimum is $35/54$ at $s=1/6$. For larger initial angles, the endpoint lifts before the hinge hits the floor, and the one-coordinate formula no longer governs the whole motion. The displayed impact speed is the intended floor-constrained answer; for unrestricted initial angles, maintained contact is an additional assumption. Subsequent free motion and possible reimpact would require a different model and contact/impact conditions.