A particle of mass m projected with a velocity ' 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 h is :
Select the correct option:
A
$\frac{\sqrt{3}}{16} \frac{m u^3}{g}$
B
$\frac{\sqrt{3}}{2} \frac{m u^2}{g}$
C
$\frac{m u^3}{\sqrt{2} g}$
D
zero
✓ Correct! Well done.
✗ Incorrect. Try again or view the solution.
Solution
$\begin{aligned} & \mathrm{L}=m u \cos \theta H \\ & =m u \cos \theta \times \frac{u^2 \sin ^2 \theta}{2 g} \\ & =\frac{m u^3}{2 g} \times \frac{\sqrt{3}}{2} \times\left(\frac{1}{2}\right)^2=\frac{\sqrt{3} m u^3}{16 g}\end{aligned}$
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