1.

The equation of projectile is y=sqrt(3)-g/2x^(2) , the angle of its projection is :

Answer»

`pi/2`
ZERO
`theta=tan^(-1)sqrt(3)`
`theta=tan^(-1)(1/sqrt(3))`

Solution :The equation of TRAJECTORY of parabolic PATH is
`y=xtantheta -(G/(2u^(2)cos^(2)theta)).x^(2)`compairing this with `y=sqrt(3x)-g x^(2)/2`
we have `tantheta=sqrt(3)`


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