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A body is projected with velocity $v_{0}$ making an angle $θ_{0}$ with the horizontal as shown in the figure. Find the angular velocity of the particle at highest point (B) relative to the point of projection (A)

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$\frac{g}{2 v_{0} \cos θ}$

$\frac{g \cos θ_{0}}{v_{0} [1 + 5 \cos^{2} θ_{0}]}$

$\frac{g}{v_{0} \sin 2 θ_{0}}$

$\frac{2 g \cos θ_{0}}{v_{0} [1 + 3 \cos^{2} θ_{0}]}$

answer is A.

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Detailed Solution

$ω_{B / A} = \frac{(v_{0} \cos θ) (\frac{H}{\sqrt{H^{2} + \frac{R^{2}}{4}}})}{\sqrt{H^{2} + \frac{R^{2}}{4}}} = \frac{v_{0} \cos θ . H}{H^{2} + \frac{R^{2}}{4}}$

Put $H = \frac{{v_{0}}^{2} \sin^{2} θ_{0}}{2 g}$ and $R = \frac{{v_{0}}^{2} \sin 2 θ_{0}}{g}$

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