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).

" f the equation of motion of a proje-ctiel is "y=3x-(1)/(8)x^(2)," the range and maximum height are respectively "(y" and "x" are in metres) "

The equation of a trajectory of a projectile is y=8x-4x^(2) . The maximum height of the projectile 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 = sqrt(3)x - ((gx^2)/2) the horizontal range is

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

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

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

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