5. In $\triangle A B C$, the median $A D$ is perpendicular to the median $B E$ at point $G$. Then the maximum value of $\sin C$ is ( ).
(A) $\frac{2}{5}$
(B) $\frac{3}{5}$
(C) $\frac{2}{3}$
(D) $\frac{4}{5}$
5. B.

In $\triangle A B C$, let $B C = a, C A = b, A B = c, G D = x, G E = y$.
Then
$$
\begin{array}{l}
G A = 2 x, \\
G B = 2 y .
\end{array}
$$

In the right triangles $\triangle A G B$, $\triangle A G E$, and $\triangle B G D$, applying the Pythagorean theorem respectively, we get
$$
4 x^{2} + 4 y^{2} = c^{2}, \quad 4 x^{2} + y^{2} = \frac{b^{2}}{4}, \quad x^{2} + 4 y^{2} = \frac{a^{2}}{4} \text{. }
$$

By the cosine rule, we have
$$
\cos C = \frac{a^{2} + b^{2} - c^{2}}{2 a b} \geqslant \frac{a^{2} + b^{2} - c^{2}}{a^{2} + b^{2}} = \frac{4}{5} \text{. }
$$

Since $\angle C$ is an acute angle, then
$$
\sin C = \sqrt{1 - \cos^{2} C} \leqslant \sqrt{1 - \left(\frac{4}{5}\right)^{2}} = \frac{3}{5} \text{. }
$$