A particle is projected from origin in xy-plane and its equation ,of trajectory is given by `y = ax - bx^(2)`. The only acceleration ,in the motion is' f' which is constant and in `-ve` direction of, y-axis.
(a) Find the velocity of projection and the angle of projection.
(b) Point of projection is considered as origin and x-axis along the horizontal ground. Find the horizontal range: and maximum height of projectile. Projectile completes its flight in horizonal plane of projection.

Explanation

Given equation of trajectory `y=ax-bx^(2)` ... ... (1)
Let u is the velocity of projection and `theta` is the angle of projection, then the equation of trajectory is given by `y=x tan thet -(f)/(2u^(2) cos^(2) theta)x^(2)` … (2)
Comparing equation (1) and (2) we get: `tan theta=a rArr theta=tan^(-1) (a) and cos theta=(1)/sqrt(1+a^(2)), f/(2u^(2) cos theta)=b`
`rArr u=1/(cos theta) sqrt((f)/(2b)) rArr u=sqrt((f)/(2b)(1+a^(2))`