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:

[JEE Main 2023, 8 Apr (Shift 2)]

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:

[JEE Main 2023, 8 Apr (Shift 2)]

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