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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, horizontal and vertical distances of the projectile from the point of projection. The maximum height attained by the particle and the angle of projectile from the horizontal are.

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, horizontal and vertical distances of the projectile from the point of projection. The maximum height attained by the particle and the angle of projectile from the horizontal are.

b are constants and x and y are horizontal and vertical distances of the projectile from the point of projection. The maximum height attained by the particle and the angle of projectile from the horizontal are. respectively The trajectory of a projectile in a vertical plane is y=ax−bx^2 Where a