The equation of the path of the projectiles is y=0.5x-0.04`x^2` .The initial speed of the projectile is in m/s (take g=10m/`s^2`)

The equation of trajectory of an oblique projectile y = sqrt(3) x - (g x^(2))/(2) . The angle of projection 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 projectile is y=16x-(x^(2))/(4) the horizontal range is:-

The equations of motion of a projectile are given by x=36t m and 2y =96t-9.8t^(2)m . The angle of projection is

Equation of trajector of ground to ground projectile is y=2x-9x^(2) . Then the angle of projection with horizontal and speed of projection is : (g=10m//s^(2))

Equation of trajector of ground to ground projectile is y=2x-9x^(2) . Then the angle of projection with horizontal and speed of projection is : (g=10m//s^(2))

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 ………………… .

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 ………………… .

A ball is projected vertically up with speed 20 m/s. Take g=10m//s^(2)

A particle of mass 1kg has been thrown with initial speed 20 m/s making an angle 60° with the horizontal ground. the angular momentum of the particle about point of projection when the projectile is at highest point is (g=10m/s^2)