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 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 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 ?

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 ?