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Physics-

General

Easy

Question

A particle of mass $m blank$ is projected with a velocity $v$ making an angle of $30 degree$ 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

$\frac{\sqrt{3}}{2} \frac{{m v}^{2}}{g}$
$Z e r o$
$\frac{{m v}^{3}}{\sqrt{2 g}}$
$\frac{\sqrt{3}}{16} \frac{{m v}^{3}}{g}$

The correct answer is: $fraction numerator square root of 3 over denominator 16 end fraction fraction numerator m v to the power of 3 end exponent over denominator g end fraction$

Angular momentum of the projectile
$L equals m v subscript h end subscript r subscript perpendicular end subscript$
$equals m open parentheses v c o s theta close parentheses h$
(where $h$ is the maximum height)
$⟹ blank equals m open parentheses v c o s theta close parentheses open parentheses fraction numerator v to the power of 2 end exponent s i n to the power of 2 end exponent theta over denominator 2 g end fraction close parentheses$
$L equals fraction numerator m v to the power of 3 end exponent s i n to the power of 2 end exponent theta c o s theta over denominator 2 g end fraction equals fraction numerator square root of 3 m v to the power of 3 end exponent over denominator 16 g end fraction$

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