The trajectory of a projectile in a vertical plane is y=ax-bx^(2), where a and b are constants and x and y are respectively horizontal and vertical distance of the projectile from the point of projection. The maximum height attained by the particle and the angle of projection from the horizontal are

The trajectory of a projectile in a vertical plane is y=ax-bx^(2), whe

1.	The trajectory of a projectile in a vertical plane is y=ax-bx^(2), where a and b are constants and x and y are respectively horizontal and vertical distance of the projectile from the point of projection. The maximum height attained by the particle and the angle of projection from the horizontal are
Answer» Solution :`y=ax-bx^(2)` For height (or y) to be MAXIMUM `(dy)/(dx)=0` or `a-2bx=0` (or) `x=a/(2b)` `:.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 `:.theta_(0)=tan^(-1)` a.