The path of a projectile is given by the equation `y = ax – bx^(2)`, where a and b are constants and x and y are respectively horizontal and vertical distances of projectile from the point of projection. The maximum height attained by the projectile and the angle of projection are respectively.

Explanation

`y = ax - bx^(2)`
For height (or y) to be maximum
`(dy)/(dX) =0 (or ) a - 2bx = 0 (or) x =(a)/(2b)`
`therefore " " y_(max) = a((a)/(2b)) - b((a)/(2b))^(2) = (a^(2))/(4b) = " "…….(ii)`
`((dy)/(dx))_(X=0) = a = tan theta_(0)` where `theta_(0)` is the angle of projection .
`therefore theta_(0) = tan^(-1) (a)`.