What is the radius of curvature of the parabola traced out by the projectile.Projected with a speed `u=sqrt30` at angle `theta=60^(@)` with the horizontal at a point where the particle velocity makes an angle `theta//2` with the horizontal ?

Answer

A

Explanation

As velocity along horizontal remains constant
`U cos theta=V cos (theta/2)`
Radius of curvature `r=v^(2)/a_(T)=(v^(2)cos^(2)theta)/(g cos^(2)(theta//2)`