The trajectory of a projectile in a vert...

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 vertical distances of the projectile from the point of projection. The maximum height attained is and the angle of projection from the horizontal is _.

`(2a^2)/(b), tan^(-1)(a)`

`(b^2)/(2a) ,tan^(-1) (b)`

`(a^2)/(b) ,tan^(-1) (2b)`

`(a^2)/(4b), tan^(-1) (a)`

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