A Particle  is projected  from the  ground  in earth’s gravitational field  at an angle   theta with  the horizontal then  Center of curvature  of projectile’s  trajectory at the highest point is below  the ground  level if  theta < tan-1 √2 / is above  the ground  level if  theta > tan-1 √2 / may be above  the ground  level if  theta < tan-1 √2./ may be above  the ground  level if  theta < tan-1 √2./  below  the ground  level if  theta > tan-1 √2.

Dear student, 

At height point of trajectory, velocity of particle is, 
v=ucosθ
If we  compare particles motion at this point with circular motion, we can write equation of motion as, 
v2r=gr=u2cos2θg
Where, r is radius of curvature.
Height of projectile is, 
H=u2sin2θ2g
If centre of curvature is below ground level, 
r<Hu2cos2θg<u2sin2θ2g2<tanθθ>tan-12If centre of curvature is above ground level,r>Hu2cos2θg>u2sin2θ2g2>tanθθ<tan-12
So, correct option is (a).

Regards.

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