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

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

Equation of trajector of ground to ground projectile is y=2x-9x^(2) . Then the angle of projection with horizontal and speed of projection is : (g=10m//s^(2))

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

The equation of the path of the projectiles is y=0.5x-0.04x^(2) .The initial speed of the projectile is? (g=10ms^(-2))

[" The equation of a trajectory of a "],[" projectile is "],[[y=3x-0.02x^(2)," where "'x'" and "'y'],[" are in meters,range of projectile is "],[" (A) "150m],[" (B) "300m],[" (C) "450m],[" (D) "600m]]