The trajectory of projectile, projected from the ground is given by...

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

$\dfrac{dy}{dx} = 0 . . . \left(1\right)$

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

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

$\begin{cases}y & = & \left(10 - \dfrac{{10}^{2}}{20}\right)m \\ & = & 5m\end{cases}$