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Easy

Question

$text A shell of mass "m' is projected with velocity end text v text at an angle end text 60 to the power of ring operator text to the horizontal. When it reaches the end text text maximum height, its angular momentum with respect to point of projection is end text$

$\frac{\sqrt{3} m v^{2}}{8 g}$
$\frac{3 m v^{3}}{8 g}$
$\frac{3 m v^{3}}{16 g}$
$\frac{\sqrt{3} m v^{3}}{16 g}$

The correct answer is: $\frac{3 m v^{3}}{16 g}$

$table attributes columnalign right left right left right left right left right left right left columnspacing 0em 2em 0em 2em 0em 2em 0em 2em 0em 2em 0em rowspacing 3 pt end attributes row cell text In vertical direction, end text 0 minus v squared sin squared invisible function application 60 to the power of ring operator equals negative 2 g h text or end text h equals fraction numerator 3 v squared over denominator 8 g end fraction end cell row cell text At highest point momentum end text equals m v cos invisible function application 60 to the power of ring operator equals fraction numerator m v over denominator 2 end fraction end cell row cell therefore text Angular momentum end text equals fraction numerator m v over denominator 2 end fraction cross times h equals fraction numerator m v over denominator 2 end fraction cross times fraction numerator 3 v squared over denominator 8 g end fraction equals fraction numerator 3 m v cubed over denominator 16 g end fraction end cell end table$

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