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.

The trajectory of a projectile is parabolic in nature. At maximum height, it can be written that

 $\frac{dy}{dx}=0 ...\left(1\right)$

Substitute the expression for trajectory into equation (1) and solve to calculate the horizontal distance, for which the projectile attains the maximum height.
$\begin{array}{rcl}\frac{d}{dx}\left[x-\frac{x^2}{20}\right]& =& 0\\ & \Rightarrow & 1-\frac{x}{10}=0\\ & \Rightarrow & x=10 \mathrm{m}\end{array}$

Substitute the value of $x$ in the given expression for the trajectory to obtain the maximum height.

$\begin{array}{rcl}y& =& \left(10-\frac{{10}^2}{20}\right) \mathrm{m}\\ & =& 5 \mathrm{m}\end{array}$