The equation of a projectile is `y=sqrt3x -(gx2)/2`,the velocity of projection is

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

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

The horizontal and vertical displacements of a projectile are given as x=at & y=bt-ct^(2) .Then velocity of projection is

The equation of trajectory of an oblique projectile y = sqrt(3) x - (g x^(2))/(2) . The angle of projection is

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

The equation of a trajectory of a projectile is y=8x-4x^(2) . The maximum height of the projectile is ?

The range of a projectile is R when the angle of projection is 40^(@) . For the same velocity of projection and range, the other possible 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 initial velocity of a projectile at the point of projection is 3hat i+4hat j ms^(-1) .Then the velocity of striking on the ground is