Question 229 of 280

A ball of mass $160g$ is thrown up at an angle of $60^{\circ}$ to the horizontal at a speed of $10m{s}^{-1}$. 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 $\left(g=10m{s}^{-2}\right)$

A$1.73kg{m}^{2}{s}^{-1}$B$3.46kg{m}^{2}{s}^{-1}$C$3.0kg{m}^{2}{s}^{-1}$D$6.0kg{m}^{2}{s}^{-1}$

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Answer: (C) $3.0kg{m}^{2}{s}^{-1}$

The angular momentum $\vec{L}=\vec{r}\times \vec{p}$
At maximum height angular momentum about $O$,
$L=mvcos\theta H$
$=mvcos\theta \times \dfrac{{u}^{2}{\sin }^{2}\theta }{2g}\left(forhorizontalv=u\right)$
$=\dfrac{m{u}^{3}{\sin }^{2}\left(60^{\circ}\right)\cos \left(60^{\circ}\right)}{2g}$
$=3kg{m}^{2}{s}^{-1}$