A shell of mass 'm' is projected with velocity V at an angle $60^{o}$ to the horizontal. When it reaches the maximum height, then its angular momentum with respect to point of projection

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Solution

The correct option is B

$\to L = m (\to v \times \to r)$
$\Rightarrow= m V r s i n 90^{o} = m V r$ [At the night point of its path]
$\Rightarrow A t h i g h e s t p o i n t V = v c o s 60^{o}; r = H = \frac{v^{2} s i n^{2} 60^{o}}{2 g}$
$\Rightarrow L = m (v c o s 60^{o}) \frac{(v s i n 60^{o})^{2}}{2 g} = \frac{3 m v^{3}}{16 g}$

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A shell of mass 'm' is projected with velocity V at an angle 60o to the horizontal. When it reaches the maximum height, then its angular momentum with respect to point of projection

The correct option is B →L=m(→v×→r) ⇒=m V r sin90o=mVr [At the night point of its path] ⇒At highest point V=vcos60o; r=H=v2sin260o2g ⇒L=m(vcos60o)(vsin60o)22g=3mv316g

A shell of mass 'm' is projected with velocity V at an angle $60^{o}$ to the horizontal. When it reaches the maximum height, then its angular momentum with respect to point of projection