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

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(B) Given: Mass $m = 160\, g = 0.16\, kg$,initial velocity $v = 10\, m/s$,angle $	heta = 60^\circ$,and acceleration due to gravity $g = 10\, m/s^2$.A

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$3.0$

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(B) Given: Mass $m = 160\, g = 0.16\, kg$,initial velocity $v = 10\, m/s$,angle $	heta = 60^\circ$,and acceleration due to gravity $g = 10\, m/s^2$.At the highest point,the vertical component of velocity is zero,and the horizontal component is $v_x = v \cos 	heta = 10 \cos 60^\circ = 10 	imes 0.5 = 5\, m/s$.The height of the highest point is $H = \frac{v^2 \sin^2 	heta}{2g} = \frac{10^2 	imes (\sin 60^\circ)^2}{2 	imes 10} = \frac{100 	imes 0.75}{20} = 3.75\, m$.The angular momentum $L$ at the highest point with respect to the point of projection is given by $L = m 	imes v_x 	imes H$.Substituting the values: $L = 0.16 	imes 5 	imes 3.75 = 0.8 	imes 3.75 = 3.0\, kg\, m^2/s$.

$A$ ball of mass $160\, g$ is thrown up at an angle of $60^\circ$ to the horizontal at a speed of $10\, m\,s^{-1}$. The angular momentum of the ball at the highest point of the trajectory with respect to the point from which the ball is thrown is nearly ........ $kg\, m^2/s$ $(g = 10\, m\,s^{-2})$.

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