5. Given a tetrahedron $S-ABC$ with the base being an equilateral triangle, point $A$'s projection $H$ on the face $SBC$ is the orthocenter of $\triangle SBC$, and $SA=a$. Then the maximum volume of this tetrahedron is $\qquad$

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5. $\frac{1}{6} a^{3}$.

From the given, $A H \perp$ plane $S B C$. Let $B H \perp S C$ at $E$. By the theorem of three perpendiculars, we know $S C \perp A E$ and $S C \perp A B$. Therefore, $S C \perp$ plane $A B E$. Draw $S O \perp$ plane $A B C$ at $O$, then $O C$ is the projection of $S C$ in plane $A B C$, so $C O \perp A B$, with the foot of the perpendicular at $F$.
Similarly, $B O \perp A C$.
Thus, point $O$ is the orthocenter of $\triangle A B C$.
Since $\triangle A B C$ is an equilateral triangle, point $O$ is also its centroid. Therefore, $S A=$
$S B=S C=a$, as shown in Figure 4.
Let $A B=x$, then $S_{\triangle A B C}$
$$
\begin{array}{l}
=\frac{\sqrt{3}}{4} x^{2}, O C=\frac{2}{3} F C= \\
- \frac{2}{3} x \sin 60^{\circ}=\frac{\sqrt{3}}{3} x, S O=\sqrt{a^{2}-\frac{1}{3} x^{2}} \text {. Therefore, } \\
V_{S-A B C}=\frac{1}{3} S_{\triangle A B C} \cdot S O=\frac{1}{12} \sqrt{x^{4}\left(3 a^{2}-x^{2}\right)} \\
=\frac{1}{6} \sqrt{\frac{1}{2} x^{2} \cdot \frac{1}{2} x^{2}\left(3 a^{2}-x^{2}\right)} \\
\leqslant \frac{1}{6} \sqrt{\left(\frac{\frac{1}{2} x^{2}+\frac{1}{2} x^{2}+3 a^{2}-x^{2}}{3}\right)^{3}}=\frac{1}{6} a^{3} \text {. } \\
\end{array}
$$

Equality holds if and only if $x=\sqrt{2} a$.
Therefore, the maximum volume of this tetrahedron is $\frac{1}{6} a^{3}$.