1.

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

Answer»

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

Solution :Here `y=usinthetaxxt-1/2g t^(2)` and
`X=ucosthetaxxt or t=x/(ucostheta)`
`y=ucostheta.x/(ucostheta)-1/2gxxx^(2)/(U^(2)cos^(2)theta)`
`=x.tantheta-1/2(GX^(2))/(u^(2)cos^(2)theta)` compairing with given equation
`y-sqrt(3)x-(gx^(2))/2` we have `tantheta=sqrt(3)`
or `theta=tan^(-1)sqrt(3)`


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