The equation of a projectile is y=3x−0.02 `x^2`. Its horizontal range is

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

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

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

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

If the equation of motion of a projectile is y=3x -1/8 x^(2) the range and maximum height are respectively (y and x are in metres)

The time of flight of a projectile is related to its horizontal range by the equation gT^2 = 2R . The angle of projection is

The equation of a projectile is y=sqrt(3)x-(gx^(2))/(2) the angle of projection is:-

The equation of motion of a projectile is y = 12 x - (3)/(4) x^2 . The horizontal component of velocity is 3 ms^-1 . What is the range of the projectile ?

The equation of motion of a projectile is y = 12 x - (3)/(4) x^2 . The horizontal component of velocity is 3 ms^-1 . What is the range of the projectile ?

A particle is projected in x-y plane with y- axis along vertical, the point of projection being origin. The equation of projectile is y = sqrt(3) x - (gx^(2))/(2) . The angle of projectile is ……………..and initial velocity is ………………… .