A stone is projected at an angle theta with v | vedclass

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(B) At the highest point of the trajectory,the vertical component of velocity is zero,and the horizontal component is $v_x = u \cos 	heta$.For a part

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

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(B) At the highest point of the trajectory,the vertical component of velocity is zero,and the horizontal component is $v_x = u \cos 	heta$.For a particle to execute circular motion,the centripetal acceleration $a_c$ is provided by the acceleration due to gravity $g$ acting perpendicular to the velocity.The formula for centripetal acceleration is $a_c = \frac{v^2}{R}$.Here,$v = v_x = u \cos 	heta$ and $a_c = g$.Substituting these values,we get $g = \frac{(u \cos 	heta)^2}{R}$.Rearranging for the radius $R$,we get $R = \frac{u^2 \cos^2 	heta}{g}$.

$A$ stone is projected at an angle $\theta$ with velocity $u$. If it executes nearly a circular motion at its maximum point for a short time,the radius of the circular path will be ($g=$ acceleration due to gravity).

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