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 ........ \(kg\, m^2/s\)  \((g\, = 10\, m\,s^{-2})\)

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