Question
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 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.