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

Options

A. `theta=tan^(-1)1/sqrt(3)`
B. `theta=tan^(-1)sqrt(3)`
C. `pi/2`
D. Zero

Answer

B

Explanation

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)`