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(B) The angular momentum is given by $\overrightarrow{L} = \overrightarrow{r} 	imes \overrightarrow{p} = \overrightarrow{r} 	imes m\overrightarrow{v

Vedclass · Accepted Answer

$\frac{\sqrt{3}}{16} \cdot \frac{mv^3}{g}$

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(B) The angular momentum is given by $\overrightarrow{L} = \overrightarrow{r} 	imes \overrightarrow{p} = \overrightarrow{r} 	imes m\overrightarrow{v}$.At the maximum height $h$,the vertical component of velocity is zero,and the velocity is purely horizontal,given by $v_x = v \cos(30^{\circ}) = \frac{\sqrt{3}v}{2}$.The position vector $\overrightarrow{r}$ at maximum height has a horizontal component equal to the horizontal range at half-time,but the cross product $\overrightarrow{r} 	imes \overrightarrow{v}$ simplifies to $L = m \cdot v_x \cdot h$.The maximum height $h$ is given by $h = \frac{v^2 \sin^2(30^{\circ})}{2g} = \frac{v^2 (1/4)}{2g} = \frac{v^2}{8g}$.Substituting these values into the angular momentum formula:$L = m \cdot \left( \frac{\sqrt{3}v}{2} ight) \cdot \left( \frac{v^2}{8g} ight) = \frac{\sqrt{3}mv^3}{16g}$.

$A$ particle of mass $m$ is projected with a velocity $v$ making an angle of $30^{\circ}$ with the horizontal. The magnitude of angular momentum of the projectile about the point of projection when the particle is at its maximum height $h$ is

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