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

Explanation

`Tantheta=arArrtheta=Tan^(-1)(a),H=a^(2)/(4b)`