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(A) The angular momentum $L$ of a particle about the point of projection is given by $L = \vec{r} 	imes \vec{p} = m(\vec{r} 	imes \vec{v})$.At the m

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$\frac{\sqrt{3}}{16} \frac{mu^3}{g}$

Vedclass · Answer

(A) The angular momentum $L$ of a particle about the point of projection is given by $L = \vec{r} 	imes \vec{p} = m(\vec{r} 	imes \vec{v})$.At the maximum height, the vertical component of velocity is zero, so the velocity is purely horizontal: $v_x = u \cos 	heta$.The horizontal distance (range) at maximum height is $x = \frac{R}{2} = \frac{u^2 \sin 	heta \cos 	heta}{g}$.The maximum height is $h = \frac{u^2 \sin^2 	heta}{2g}$.The angular momentum is $L = m v_x h = m (u \cos 	heta) \left( \frac{u^2 \sin^2 	heta}{2g} ight)$.Substituting $	heta = 30^{\circ}$, $\cos 30^{\circ} = \frac{\sqrt{3}}{2}$, and $\sin 30^{\circ} = \frac{1}{2}$:$L = m u \left( \frac{\sqrt{3}}{2} ight) \frac{u^2 (1/2)^2}{2g} = m u \left( \frac{\sqrt{3}}{2} ight) \frac{u^2}{8g} = \frac{\sqrt{3} m u^3}{16g}$.

$A$ particle of mass $m$ is projected with a velocity $u$ making an angle of $30^{\circ}$ 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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