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A particle of mass '1 \mathrm{~kg} ' is projected with a velocity of ' 10 \mathrm{~m} / \mathrm{s} ' at an angle of 30^{\circ} with the horizontal. Then the magnitude of its angular momentum, just before it hits the ground will be !- [Take, \left.g=10 \mathrm{~m} / \mathrm{s}^2\right]

Option: 1

25 \mathrm{~kg}-\mathrm{m}^2-\mathrm{s}^{-1}


Option: 2

25 \sqrt{3} \mathrm{~kg}-\mathrm{m}^2-\mathrm{s}^{-1}


Option: 3

75 \mathrm{~kg}-\mathrm{m}^2-\mathrm{s}^{-1}


Option: 4

75 \sqrt{3} \mathrm{~kg}-\mathrm{m}^2-\mathrm{s}^{-1}


Answers (1)

best_answer

Angular momentum of the particle at 'B'

is given by

\begin{aligned} & L=m(\vec{r} \times \vec{v}) \quad \\ & L=m R u \sin \theta \\ & L=(1)\left[\frac{(10)^2 \sin 2 \times 30^{\circ}}{10}\right] \times 10 \times \sin 30^{\circ} \\ & L=10 \times \frac{\sqrt{3}}{2} \times 10 \times \frac{1}{2} \\ & L=25 \sqrt{3} \mathrm{~kg}-\frac{m^2}{\mathrm{~s}} \end{aligned}

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Sayak

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