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.

We have the trajectory of the projectile given by

$$y = x - \frac{x^2}{20}$$

To find the maximum height, we set $$\frac{dy}{dx} = 0$$:

$$\frac{dy}{dx} = 1 - \frac{2x}{20} = 1 - \frac{x}{10} = 0$$

Solving gives $$x = 10$$ m. Now substituting $$x = 10$$ back into the trajectory equation:

$$y_{max} = 10 - \frac{(10)^2}{20} = 10 - \frac{100}{20} = 10 - 5 = 5 \text{ m}$$

So the maximum height attained by the projectile is 5 m. Hence, the correct answer is Option 3.