A particle of mass m is projected with veloci | vedclass

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(B) At the highest point,the vertical component of velocity is zero,and the horizontal component is $v_x = v \cos 45^{\circ} = v / \sqrt{2}$.The momen

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$mv^3 / (4\sqrt{2}g)$

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(B) At the highest point,the vertical component of velocity is zero,and the horizontal component is $v_x = v \cos 45^{\circ} = v / \sqrt{2}$.The momentum of the particle at this point is $p = m v_x = mv / \sqrt{2}$.The angular momentum $L$ about the point of projection is given by $L = p 	imes h$,where $h$ is the maximum height.The maximum height $h$ is given by $h = (v^2 \sin^2 45^{\circ}) / (2g) = (v^2 \cdot (1/2)) / (2g) = v^2 / (4g)$.Substituting these values,we get $L = (mv / \sqrt{2}) 	imes (v^2 / 4g) = mv^3 / (4\sqrt{2}g)$.

$A$ particle of mass $m$ is projected with velocity $v$ making an angle of $45^{\circ}$ with the horizontal. The magnitude of the angular momentum of the particle about the point of projection when the particle is at its maximum height is (where $g$ is the acceleration due to gravity):

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