A ball of mass 160 , g is thrown up at an ang | vedclass

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(B) The angular momentum of a particle is given by $\vec{L} = \vec{r} 	imes \vec{p} = \vec{r} 	imes m\vec{v}$.At the highest point,the velocity of t

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(B) The angular momentum of a particle is given by $\vec{L} = \vec{r} 	imes \vec{p} = \vec{r} 	imes m\vec{v}$.At the highest point,the velocity of the projectile is purely horizontal,given by $v_x = v \cos 	heta$.The horizontal distance (range) at the highest point is $x = R/2 = \frac{v^2 \sin 	heta \cos 	heta}{g}$.The vertical height at the highest point is $H = \frac{v^2 \sin^2 	heta}{2g}$.The angular momentum about the point of projection is $L = m v_x H = m (v \cos 	heta) \left( \frac{v^2 \sin^2 	heta}{2g} ight) = \frac{m v^3 \sin^2 	heta \cos 	heta}{2g}$.Given: $m = 0.16 \, kg$,$v = 10 \, m/s$,$	heta = 60^{\circ}$,$g = 10 \, m/s^2$.$L = \frac{0.16 	imes (10)^3 	imes \sin^2 60^{\circ} 	imes \cos 60^{\circ}}{2 	imes 10}$.$L = \frac{0.16 	imes 1000 	imes (3/4) 	imes (1/2)}{20} = \frac{160 	imes 0.375}{20} = 8 	imes 0.375 = 3.0 \, kg \cdot m^2/s$.

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