The path of a projectile is given by the equation y=a x-b x^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

Step-by-step solution

The given equation, y=a x-b x^2 and we know that equation of trajectory is y=( ) x- 1 2 g u^2 ^2 x^2 Compare both equations, we get a= , b= 1 2 g u^2 ^2 The maximum height aligned a^2 b & = ^2 g 2 u^2 ^2 & = ^2 g ^2 2 u^2 ^2 & = 2 u^2 ^2 g =4 ( u^2 ^2 2 g ) H_ & = a^2 4 b a & = _ & = ⁻¹(a) aligned and aligned & a= & = ⁻¹(a) aligned So, required solution is a^2 4 b , ⁻¹(a)