The trajectory of projectile, projected from the ground is given by $y=x-\frac{x^{2}}{20}$. Where $x$ and $y$ are measured in meter. The maximum height attained by the projectile will be.

<p>The equation of the trajectory given is $y = x - \frac{x^2}{20}$.<br/><br/> This is a parabola, and it represents the path of the projectile.</p> <p>The maximum height of the projectile corresponds to the vertex of the parabola.<br/><br/> The x-coordinate of the vertex for a parabola given by $y = ax^2 + bx + c$ is $-b/2a$.<br/><br/> In this case, $a = -1/20$ and $b = 1$, so the x-coordinate of the vertex is:</p> <p>$ x_{\text{vertex}} = -\frac{b}{2a} = -\frac{1}{2 \times (-1/20)} = 10 $</p> <p>Substituting this into the equation of the trajectory gives the y-coordinate of the vertex, which is the maximum height:</p> <p>$ y_{\text{max}} = 10 - \frac{10^2}{20} = 10 - 5 = 5 $</p> <p>So, the maximum height attained by the projectile is 5 m.</p>