A particle of mass m is projected with speed v at an angle of 30° with the horizontal, find its angular momentum about point of projection when it reaches its maximum height.

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Projectile Motion Physics Motion in Plane Physics Class 11th

A particle of mass m is projected with speed v at an angle of 30° with the horizontal, find its angular momentum about point of projection when it reaches its maximum height.

Option 1 -

$\frac{m v^{3}}{16 g}$

Option 2 -

$\sqrt[]{3} \frac{m v^{3}}{16 g}$

Option 3 -

$\frac{m v^{3}}{3 g}$

Option 4 -

$\sqrt[]{3} \frac{m v^{3}}{8 g}$

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1 Answer

Answered by

alok kumar singh | Contributor-Level 10

a month ago

Correct Option - 2

Detailed Solution:

Velocity at maximum height = vcoss30°

L = m (vcos30) H

$= m v (\frac{\sqrt[]{3}}{2}) * \frac{v^{2} s i n^{2} 30}{2 g}$

$= \sqrt[]{3} \frac{m v^{3}}{16 g}$

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