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(D) The radius of curvature $R$ of a trajectory is given by $R = \frac{v^2}{a_{\perp}}$,where $v$ is the speed and $a_{\perp}$ is the component of acc

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$\frac{1}{\cos ^3 	heta}$

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(D) The radius of curvature $R$ of a trajectory is given by $R = \frac{v^2}{a_{\perp}}$,where $v$ is the speed and $a_{\perp}$ is the component of acceleration perpendicular to the velocity.At the point of projection $(A)$:The speed is $u$. The acceleration due to gravity $g$ acts vertically downwards. The angle between the velocity vector and the vertical is $(90^\circ - 	heta)$. Thus,the component of $g$ perpendicular to the velocity is $g \cos 	heta$.Therefore,$r_A = \frac{u^2}{g \cos 	heta}$.At the maximum height $(H)$:The speed is $v_H = u \cos 	heta$. The acceleration due to gravity $g$ acts vertically downwards,which is perpendicular to the horizontal velocity at this point.Therefore,$r_H = \frac{(u \cos 	heta)^2}{g} = \frac{u^2 \cos ^2 	heta}{g}$.The ratio of the radius of curvature at the point of projection to the radius of curvature at the maximum height is:$\frac{r_A}{r_H} = \frac{\frac{u^2}{g \cos 	heta}}{\frac{u^2 \cos ^2 	heta}{g}} = \frac{1}{\cos 	heta} \cdot \frac{1}{\cos ^2 	heta} = \frac{1}{\cos ^3 	heta}$.

$A$ particle is projected from the ground at an angle $\theta$ with the horizontal with speed $u$. The ratio of the radius of curvature of its trajectory at the point of projection to the radius of curvature at the maximum height is ........

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