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(A) The particle is projected with speed $u = 10 \ m/s$ at an angle $	heta = 60^\circ$ with the horizontal.Due to the symmetry of projectile motion,w

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$20$

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(A) The particle is projected with speed $u = 10 \ m/s$ at an angle $	heta = 60^\circ$ with the horizontal.Due to the symmetry of projectile motion,when the particle returns to the same horizontal level,its speed remains $u = 10 \ m/s$,and the angle it makes with the horizontal is also $	heta = 60^\circ$,but directed downwards.The radius of curvature $R$ is given by the formula $R = \frac{v^2}{a_n}$,where $v$ is the speed and $a_n$ is the normal acceleration.The normal acceleration $a_n$ is the component of acceleration perpendicular to the velocity vector.At the point of landing,the acceleration due to gravity $g$ acts vertically downwards.The angle between the velocity vector and the horizontal is $60^\circ$,so the angle between the velocity vector and the vertical is $30^\circ$.The component of $g$ perpendicular to the velocity is $a_n = g \cos(60^\circ)$.Substituting the values: $R = \frac{u^2}{g \cos(60^\circ)} = \frac{10^2}{10 	imes 0.5} = \frac{100}{5} = 20 \ m$.

$A$ heavy particle is projected from a point on the horizontal at an angle $60^\circ$ with the horizontal with a speed of $10 \ m/s$. Then the radius of curvature of its path at the instant of crossing the same horizontal will be ......... $m$.

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