The equation of trajectory of a projectile is given by y = x - 5x^2...

The equation of the trajectory is $y = x - 5x^2$.

The maximum height corresponds to the maximum value of the vertical displacement $y$. To find the maximum value of $y$, we differentiate $y$ with respect to $x$ and set it to zero:
$\dfrac{dy}{dx} = 1 - 10x = 0$
$\implies 10x = 1 \implies x = 0.1 \mathrm{~m}$

This means the projectile reaches its maximum height at a horizontal distance of $x = 0.1 \mathrm{~m}$.

To find the maximum height, substitute $x = 0.1 \mathrm{~m}$ back into the trajectory equation:
$y_{\text{max}} = (0.1) - 5(0.1)^2$
$y_{\text{max}} = 0.1 - 5(0.01)$
$y_{\text{max}} = 0.1 - 0.05 = 0.05 \mathrm{~m}$

Alternatively, comparing $y = x - 5x^2$ with the standard trajectory equation $y = x \tan\theta - \dfrac{gx^2}{2u^2 \cos^2\theta}$, we get $\tan\theta = 1 \implies \theta = 45^{\circ}$.
The second term gives $\dfrac{g}{2u^2 \cos^2(45^{\circ})} = 5 \implies \dfrac{10}{2u^2 (1/2)} = 5 \implies \dfrac{10}{u^2} = 5 \implies u^2 = 2$.
Then, maximum height $H = \dfrac{u^2 \sin^2\theta}{2g} = \dfrac{2 \times (1/\sqrt{2})^2}{2 \times 10} = \dfrac{1}{20} = 0.05 \mathrm{~m}$.

Answer: $0.05 \mathrm{~m}$