The trajectory of a projectile in a vertical plane is given by the equation, y=ax-bx2,  where a and b are constants 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 projection with respect to the horizontal are

given equation =y=ax−bx2---(1)    when y is maximum dydx   is zero  to find maximum value of y, differentiate y with respect to x and equate to zero,dydx=a−2bxequate dydx=0a−2bx=0 x=a2b---(2)substitute eqn(2) in eqn(1),∴ymax=aa2b−ba2b2=a24b=maximum heightto find angle of projection,  compare  y=ax-bx2 with y=tanθ x-gx22 u2cos2θ ⇒tanθ=a      where θ is the angle ofprojectionθ=tan−1(a)=angle of projection