A particle of mass $\mathrm{m}$ is projected with a velocity '$\mathrm{u}$' making an angle of $30^{\circ}$ with the horizontal. The magnitude of angular momentum of the projectile about the point of projection when the particle is at its maximum height $\mathrm{h}$ is :
Solution
<p>$$\begin{aligned}
& \mathrm{L}=m u \cos \theta H \\
& =m u \cos \theta \times \frac{u^2 \sin ^2 \theta}{2 g} \\
& =\frac{m u^3}{2 g} \times \frac{\sqrt{3}}{2} \times\left(\frac{1}{2}\right)^2=\frac{\sqrt{3} m u^3}{16 g}
\end{aligned}$$</p>
About this question
Subject: Physics · Chapter: Rotational Motion · Topic: Centre of Mass
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