[" Jhe parabelic path of a projectiberepetentid by "],[qquad y=(x)/(sqrt(3))-(x^(2))/(6^(@))" in MKS units.It angle of "],[" prigedion is "(9=10ms^(-2))]

The parabolic path of a projectile is represented by y=x/sqrt3-x^(2)/60 in MKS units: Its angle of projection is (g=10ms^(-2))

The angle between y=x+6 and y=sqrt(3)x+y is

The angle of rotation of axes to remove xy terms in the equation 3x^(2) - 2 sqrt(3) xy + 9y^(2) = 10 is

Area common to the curve y=sqrt(9-x^(2)) and x^(2)+y^(2)=6x is

Angle between y = x + 6 and y = sqrt(3)x + 7 is

What will be the equation for the path of a projectile if it is fired with initial velocity of (2hati+3hatj) . Here hatu and hatj are unit vectors along X-axis and Y-axis (g=10ms^(-2)) .

The angle of rotation of axes to remove xy term in the equation 9x^(2) + 2sqrt(3)xy + 7y^(2)=10 is

The angle of rotation of axes to remove xy term in the equation 9x^(2) + 2sqrt(3)xy + 7y^(2)=10 is

Solve : (5+2sqrt6)^(x^(2)-3) +(5-2sqrt6)^(x^(2)-3) = 10