[" The equation of trajectory of a projectile is "],[y=ax-bx^(2)" where "x" and "y" axes are along "],[" horizontal and vertical directions "],[" respectively.The range of the projectile is "]

The equation of a trajectory of a projectile is y=8x-4x^(2) . The maximum height of the projectile is ?

The initial velocity of a projectile is (hati+2hatj)m//s , where hati and hatj are unit vectors along the horizontal and vertical directions respectively. Find out the locus of the projectile , taking g=10 m//s^2 .

The equation of motion of a projectile is y =ax - bx^2 where a and b are constants of motion. The horizontal range of the projectile is

The equation of a projectile is y = ax - bx^(2) . Its horizontal range is

The equation of a projectile is y = ax - bx^(2) . Its horizontal range is

The equation of path of a particle projected obliquely from the ground under gravity is given by y=2x-x^(2) where 'x' and ' y' are in meter.The "x" and "y" axes are in horizontal and vertical directions respectively.The speed of projection is

The equation of the trajectory of a projectile on a vertical plane is y= ax-bx^2 , where a and b are constants, and x and y respectively are the horizontal distances of the projectile from the point of projection. Find out the maximum height attained by the projectile, and the angle of projection with respect to the horizontal.

The equation of trajectory of a projectile is y=10x-(5/9)x^2 if we assume g=10ms^(-3) then range of projectile (in metre) is