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(B) At the maximum height,the vertical component of velocity is zero,and the horizontal component of velocity is $v = u \cos 	heta$.The acceleration

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

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(B) At the maximum height,the vertical component of velocity is zero,and the horizontal component of velocity is $v = u \cos 	heta$.The acceleration acting on the object at this point is the acceleration due to gravity,$g$,which acts vertically downwards.The centripetal acceleration $a_c$ is given by $a_c = \frac{v^2}{R}$,where $R$ is the radius of curvature.At the maximum height,the acceleration $g$ is perpendicular to the velocity $v$. Thus,$g$ acts as the centripetal acceleration.Therefore,$g = \frac{(u \cos 	heta)^2}{R}$.Rearranging for $R$,we get $R = \frac{u^2 \cos^2 	heta}{g}$.

$P$. Time of flight	$1$. $3$
$Q$. Time when particle is moving at an angle of $53^{\circ}$ with horizontal in upwards direction	$2$. $10$
$R$. Range of the particle	$3$. $50$
$S$. Maximum height reached by the particle	$4$. $60$

An object is projected from the ground with speed $u$ at an angle $\theta$ with the horizontal. The radius of curvature of its trajectory at the maximum height from the ground is ..........

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