A particle is projected from the ground in earth’s gravitational field at an angle $θ$ with the horizontal then

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center of curvature of projectile’s trajectory at the highest point is below the ground level if $θ < \tan^{- 1} \sqrt{2}$

Center of curvature of projectile’s trajectory at the highest point is above the ground level if $θ > \tan^{- 1} \sqrt{2}$

Center of curvature of projectile’s trajectory at the highest point is above ground level if $θ < \tan^{- 1} 2$

Center of curvature of projectile’s trajectory at the highest point is below the ground level if $θ > \tan^{- 1} \sqrt{2}$

answer is A, B.

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Detailed Solution

If $\frac{u^{2} \cos^{2} θ}{g} > \frac{u^{2} \sin^{2} θ}{2 g}$

$θ < \tan^{- 1} \sqrt{2}$

If $\frac{u^{2} \cos^{2} θ}{g} < \frac{u^{2} \sin^{2} θ}{2 g}$

$θ > \tan^{- 1} \sqrt{2}$

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