Let $l$ be the given slant height of the cone, $r$ be the radius of the base, $h$ be the height, and $\alpha$ be the semi-vertical angle.

From the geometry of the cone, we have:

$$r = l \sin\alpha \quad \text{and} \quad h = l \cos\alpha$$

Volume of the cone:

$$V = \frac{1}{3}\pi r^2 h$$

Substituting $r$ and $h$ in terms of $\alpha$:

$$V = \frac{1}{3}\pi (l \sin\alpha)^2 \cdot l \cos\alpha$$ $$V = \frac{1}{3}\pi l^3 \sin^2\alpha \cos\alpha$$

Since $l$ is constant, to maximize $V$, we need to maximize $\sin^2\alpha \cos\alpha$.

Differentiating with respect to $\alpha$:

$$\frac{dV}{d\alpha} = \frac{1}{3}\pi l^3 \cdot \frac{d}{d\alpha}(\sin^2\alpha \cos\alpha)$$

Using product rule:

$$\frac{d}{d\alpha}(\sin^2\alpha \cos\alpha) = 2\sin\alpha \cos\alpha \cdot \cos\alpha + \sin^2\alpha \cdot (-\sin\alpha)$$ $$= 2\sin\alpha \cos^2\alpha - \sin^3\alpha$$ $$= \sin\alpha(2\cos^2\alpha - \sin^2\alpha)$$

For critical points:

$$\frac{dV}{d\alpha} = 0$$

Since $\sin\alpha \neq 0$ for a valid cone:

$$2\cos^2\alpha - \sin^2\alpha = 0$$ $$2\cos^2\alpha = \sin^2\alpha$$ $$\tan^2\alpha = 2$$ $$\tan\alpha = \sqrt{2}$$

Therefore:

$$\alpha = \tan^{-1}\sqrt{2}$$

Second derivative test:

At $\tan\alpha = \sqrt{2}$, we have $\sin\alpha = \frac{\sqrt{2}}{\sqrt{3}}$ and $\cos\alpha = \frac{1}{\sqrt{3}}$.

Computing $\frac{d^2V}{d\alpha^2}$ and substituting these values gives a negative result, confirming maximum volume.

∴ Hence, the semi-vertical angle of the cone of maximum volume with given slant height is $\tan^{-1}\sqrt{2}$.