Question Answers

Sahay Sir > Question Answers > Uncategorised (JEE Advanced Physics by BM Sharma + GMP Solutions) > The trajectory of a projectile in a vertical plane is y = ax – bx^2, where a, b are constants, and x and y are, respectively, the horizontal and vetical distances of the projectile from the point of projection.

The trajectory of a projectile in a vertical plane is y = ax – bx^2, where a, b are constants, and x and y are, respectively, the horizontal and vetical distances of the projectile from the point of projection.

and x and y are b are constants respectively the horizontal and vetical distances of the projectile from the point of projection. The trajectory of a projectile in a vertical plane is y = ax - bx^2 Where a