A particle is projected from the origin in X-Y plane. Acceleration of particle in Y direction is a. If equation of path of the particle is `y = ax-bx^(2)`, then find initial velocity of the particle.

Explanation

`y=ax- bx^(2) " " y= x tan theta= (alpha x^(2))/(2 u cos^(2) theta)`
`tan theta= a and (alpha)/(2u cos^(2) theta)=b u= sqrt((alpha(1+a^(2)))/(2b)`