A ball of mass 160g is thrown up at an angle of $60 °$ to the horizontal at a speed of $10 {ms}^{- 1}$ . 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 $(g = 10 {ms}^{- 2})$

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$3.46 kg m^{2} / s$

$6.0 kg m^{2} / s$

$3.0 kg m^{2} / s$

$1.73 kg m^{2} / s$

answer is B.

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Detailed Solution

$\begin{array}{l} Angular momentum L = \vec{r} \times m \vec{v} \\ = Hmv \cos θ \\ = \frac{v^{2} \sin^{2} θ}{2 g} m v \cos θ [H = \frac{v^{2} \sin^{2} θ}{2 g}] \\ = \frac{10^{2} \times \sin^{2} 60^{°} \times 0.160 \times 10 \times \cos 60^{°}}{2 \times 10} \\ = 3 kg m^{2} / s \end{array}$