Cannon

The trajectory of the projectile from the point of firing is $y \; = \; x\tan\theta - \frac{gx^2}{2V^2\cos^2\theta}$ where $V$ is the muzzle velocity and $\theta$ the angle of elevation. Since the projectile must pass through the point $(\tfrac12D\sqrt{3},-\tfrac12D)$ , we can show that $V^2 \; = \; \frac{3gD \sec^2\theta}{4(1+\sqrt{3}\tan\theta)}$ and since $\frac{d}{d\theta}V^2 \; = \; \frac{3gD\sec^2\theta}{4(1+\sqrt{3}\tan\theta)^2}(\sqrt{3}\tan\theta - 1)(\tan\theta+\sqrt{3})$ the turning point of $V^2$ (which is in fact the minimum) occurs when $\tan\theta = \frac{1}{\sqrt{3}}$ . Substituting in, the minimum value of $V$ is $\sqrt{\tfrac12gD}$ .

With $g=10$ and $D=6000$ , we obtain $V_\text{min} = \boxed{173.205..}$ .