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If a particle is projected with speed u from the ground at an angle $θ$ with horizontal. Radius of curvature of a point where velocity vector is perpendicular to initial velocity vector is given by

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a

$\frac{u^{2} \cot^{2} θ}{g \sin θ}$

b

$\frac{u^{2} \tan^{2} θ}{g \cos θ}$

c

$\frac{u^{2}}{g}$

d

$\frac{u^{2} \cos^{2} θ}{g}$

answer is B.

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Detailed Solution

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$v \cos (90^{0} - θ) = u \cos θ o r v \sin θ = u \cos θ; v = u \cot θ$

$\frac{v_{T}^{2}}{R} = a_{c}; \frac{u^{2} \cot^{2} θ}{g \sin θ} = R$

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If a particle is projected with speed u from the ground at an angle θ with horizontal. Radius of curvature of a point where velocity vector is perpendicular to initial velocity vector is given by