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

Answer

C