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Physics Question on Uniform Circular Motion

A body is projected with a speed 'u? at an angle ′θ′'\theta' with the horizontal. The radius of curvature of the trajectory when it makes an angle (θ2)\left(\frac{\theta}{2} \right) with the horizontal is (g - acceleration due to gravity)

A

u2cos⁡2θsec⁡3(θ/2)3g\frac{u^{2} \cos^{2} \theta \sec^{3} \left( \theta/2\right)}{\sqrt{3}g}

B

u2cos⁡2θsec⁡3(θ/2)2g\frac{u^{2} \cos^{2} \theta \sec^{3} \left( \theta/2\right)}{2 g}

C

2u2cos⁡2θsec⁡3(θ/2)g\frac{ 2 u^{2} \cos^{2} \theta \sec^{3} \left( \theta/2\right)}{g}

D

u2cos⁡2θsec⁡3(θ/2)g\frac{u^{2} \cos^{2} \theta \sec^{3} \left( \theta/2\right)}{g}

Answer

u2cos⁡2θsec⁡3(θ/2)g\frac{u^{2} \cos^{2} \theta \sec^{3} \left( \theta/2\right)}{g}

Explanation

Solution

Answer (d) u2cos⁡2θsec⁡3(θ/2)g\frac{u^{2} \cos^{2} \theta \sec^{3} \left( \theta/2\right)}{g}