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

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Question

$\begin{array}{l}
Given\,:\,m = 0.160kg\
	heta  = {60^ \circ }\
V = 10\,m/s\
Angular\,momentu\,m\,L = \vec r 	imes m\,\vec V\
 =

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Vedclass · Answer

$\begin{array}{l}
Given\,:\,m = 0.160kg\
	heta  = {60^ \circ }\
V = 10\,m/s\
Angular\,momentu\,m\,L = \vec r 	imes m\,\vec V\
 = H\,mv\,\cos 	heta \
 = \,mv\frac{{{V^2}{{\sin }^2}	heta }}{{2g}}\cos 	heta \,\,\,\left[ {H = \frac{{{V^2}{{\sin }^2}	heta }}{{2g}}} ight]\
 = \frac{{{{10}^2} 	imes {{\sin }^2}{{60}^ \circ } 	imes \cos {{60}^ \circ }}}{{2 	imes 10}} 	imes \left( {0.16 	imes 10} ight)\
 = 3\,kg\,{m^2}/s
\end{array}$

A ball of mass $160\, g$ is thrown up at an angle of $60^o$ to the horizontal at a speed of $10\, m\,s^{-1}$ . The angular momentum of the ball at the highest point of the trajectcry 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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