The trajectory of a projectile in a vertical plane is `y=sqrt(3)x-2x^(2)`. `[g=10 m//s^(2)]`
Channel: Doubtnut
Date: 2020-04-12

The trajectory of a projectile in a vertical plane is `y=sqrt(3)x-2x^(2)`. `[g=10 m//s^(2)]`   mg src="https://d10lpgp6xz60nq.cloudfront.net/physics_images/ALN_PHY_C02_E01_150_Q01.png" width="80%"   Radius of curvature of the path of the projectile at the topmost point P is :-