A body of mass 'm' is projected with a speed | vedclass

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

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$8$

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(A) The angular momentum $L$ about the point of projection is given by $L = \vec{r} 	imes \vec{p} = m(\vec{r} 	imes \vec{v})$.At the highest point, the velocity is $v_x = u \cos 45^{\circ} = \frac{u}{\sqrt{2}}$ and the vertical height is $H = \frac{u^2 \sin^2 45^{\circ}}{2g} = \frac{u^2}{4g}$.The angular momentum is $L = m v_x H = m \left( \frac{u}{\sqrt{2}} ight) \left( \frac{u^2}{4g} ight) = \frac{m u^3}{4\sqrt{2} g}$.To match the form $\frac{\sqrt{2} m u^3}{Xg}$, we multiply the numerator and denominator by $\sqrt{2}$:$L = \frac{m u^3 \sqrt{2}}{4 \sqrt{2} \cdot \sqrt{2} g} = \frac{\sqrt{2} m u^3}{8g}$.Comparing this with $\frac{\sqrt{2} m u^3}{Xg}$, we get $X = 8$.

$A$ body of mass '$m$' is projected with a speed '$u$' making an angle of $45^{\circ}$ with the ground. The angular momentum of the body about the point of projection, at the highest point is expressed as $\frac{\sqrt{2} mu^3}{Xg}$. The value of '$X$' is

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