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

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The trajectory of a projectile is parabolic in nature. At maximum height, it can be written that
$\frac{d y}{d x} = 0 . . . (1)$
Substitute the expression for trajectory into equation (1) and solve to calculate the horizontal distance, for which the projectile attains the maximum height.
$\begin{array}{rcl} \frac{d}{d x} [x - \frac{x^{2}}{20}] & = & 0 \\ \Rightarrow & 1 - \frac{x}{10} = 0 \\ \Rightarrow & x = 10 m \end{array}$
Substitute the value of $x$ in the given expression for the trajectory to obtain the maximum height.
$\begin{array}{rcl} y & = & (10 - \frac{10^{2}}{20}) m \\ = & 5 m \end{array}$

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