The equation of trajectory ofa projectile is `y = 10x-5/9x^2` If we assume `g = 10 ms^(-2)`, the range of projectile (in metre) is

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

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

The equation of trajectory of an oblique projectile y = sqrt(3) x - (g x^(2))/(2) . The angle of projection is

The equation of trajectory of an oblique projectile y = sqrt(3) x - (g x^(2))/(2) . The angle of projection is

The equation of a trajectory of a projectile is y=8x-4x^(2) The maximum height of the projectile is (A) 8m (B) 6m (C) 2m (D) 4m

The equation of projectile is y = 16 x - (5)/(4)x^(2) find the horizontal range

The equation of projectile is y = 16 x - (5)/(4)x^(2) find the horizontal range

[" 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 "]