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(D) At the maximum height $h$,the vertical component of velocity is zero,and the horizontal component of velocity is $V_x = V \cos 45^{\circ} = \frac{

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$m\sqrt{2gh^3}$

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(D) At the maximum height $h$,the vertical component of velocity is zero,and the horizontal component of velocity is $V_x = V \cos 45^{\circ} = \frac{V}{\sqrt{2}}$.The momentum of the particle at the maximum height is $p = m V_x = \frac{mV}{\sqrt{2}}$.The angular momentum $L$ about the point of projection is given by $L = p 	imes h$,where $h$ is the perpendicular distance (maximum height) from the point of projection.The maximum height $h$ is given by $h = \frac{V^2 \sin^2 45^{\circ}}{2g} = \frac{V^2}{4g}$.Substituting these into the angular momentum formula:$L = \left( \frac{mV}{\sqrt{2}} ight) 	imes h = \frac{mV}{\sqrt{2}} 	imes \frac{V^2}{4g} = \frac{mV^3}{4\sqrt{2}g}$.Since $h = \frac{V^2}{4g}$,we have $V^2 = 4gh$,which implies $V = \sqrt{4gh} = 2\sqrt{gh}$.Substituting $V$ into the expression for $L$:$L = \frac{m(2\sqrt{gh})^3}{4\sqrt{2}g} = \frac{m(8g^{3/2}h^{3/2})}{4\sqrt{2}g} = \frac{2m g h \sqrt{gh}}{\sqrt{2}g} = \frac{2}{\sqrt{2}} m \sqrt{gh^3} = m\sqrt{2gh^3}$.

$A$ particle of mass $m$ is projected with a velocity $V$ making an angle of $45^{\circ}$ with the horizontal. The magnitude of the angular momentum of the projectile about the point of projection when the particle is at its maximum height $h$ is

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