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A particle is projected horizontally from the top of a tower with a velocity $v_{0}$ . If $v$ be its velocity at any instant, then the radius of curvature of the path of the particle at that instant is directly proportional to :

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$\frac{1}{v}$

$v^{3}$

$v^{2}$

$v$

answer is A.

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Detailed Solution

As we know : $a_{c} = \frac{v^{2}}{R}$ (Centripetal acceleration)

From figure, $g \sin θ = \frac{V^{2}}{R}$

$\Rightarrow g \cdot \frac{v_{0}}{V} = \frac{v^{2}}{R} (∵ \sin θ_{i} = \frac{v_{0}}{V}) \Rightarrow R \propto v^{3}$

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