The equation of the trajectory given is $y = x - \frac{x^2}{20}$.

This is a parabola, and it represents the path of the projectile.

The maximum height of the projectile corresponds to the vertex of the parabola.

The x-coordinate of the vertex for a parabola given by $y = ax^2 + bx + c$ is $-b/2a$.

In this case, $a = -1/20$ and $b = 1$, so the x-coordinate of the vertex is:

$ x_{\text{vertex}} = -\frac{b}{2a} = -\frac{1}{2 \times (-1/20)} = 10 $

Substituting this into the equation of the trajectory gives the y-coordinate of the vertex, which is the maximum height:

$ y_{\text{max}} = 10 - \frac{10^2}{20} = 10 - 5 = 5 $

So, the maximum height attained by the projectile is 5 m.