The trajectory of a projectile in a vertical plane is `y = ax - bx^2`, where `a and b` are constant 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.

Options

A. `(2a^2)/(b), tan^(-1) (a)`
B. `(b^2)/(2a), tan^(-1) (b) `
C. `(a^2)/(b), tan^(-1) (2b)`
D. `(a^2)/(4b) , tan^(-1) (a)`

Answer

D