For ground to ground projectile motion equation of path is `y=12x-3//4x^(2)`. Given that `g=10ms^(-2)`. 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 ?

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 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 trajectory ofa projectile is y = 10x-5/9x^2 If we assume g = 10 ms^(-2) , the range of projectile (in metre) is

The equation of trajectory ofa projectile is y = 10x-5/9x^2 If we assume g = 10 ms^(-2) , the 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 trajectory of a projectile in a vertical plane is y=sqrt(3)x-2x^(2) . [g=10 m//s^(2)] Radius of curvature of the path of the projectile at the topmost point P is :-

The trajectory of a projectile in a vertical plane is y=sqrt(3)x-2x^(2) . [g=10 m//s^(2)] Radius of curvature of the path of the projectile at the topmost point P is :-

The trajectory of a projectile in a vertical plane is y=sqrt(3)x-2x^(2) . [g=10 m//s^(2)] Time of flight of the projectile is :-