6. As shown in Figure 6, the maximum height of a parabolic arc is 15, and the span is 60. Then, the height of the arc at a point 12 units away from the midpoint $M$ is $\qquad$.
6. $12 \frac{3}{5}$.

With $M$ as the origin, the line $AB$ as the $x$-axis, and the line $MC$ as the $y$-axis, establish a Cartesian coordinate system. The parabola is the graph of the quadratic function $y = a x^{2} + 15$.
Therefore, point $B(30,0)$ lies on the parabola, which means $900 a + 15 = 0$.
Thus, $a = -\frac{1}{60}, y = -\frac{1}{60} x^{2} + 15$.
When $x = 12$, $y = -\frac{1}{60} \times 144 + 15 = 12 \frac{3}{5}$.