[" The equation of projectile is "y=16x-(x^(2))/(4)" the "],[" horizontal range is ":-],[[" (1) "16m," (2) "8m," (3) "64m," (4) "128m]]

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

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

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 a projectile is y = sqrt(3)x - ((gx^2)/2) the horizontal range is

The equation of a projectile is y = sqrt(3)x - ((gx^2)/2) the horizontal range 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 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]]

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