A particle of mass $m$ is projected with velocity $v$ making an angle of ${45^o}$with the horizontal. The magnitude of the angular momentum of the particle about the point of projection when the particle is at its maximum height is (where $g = $ acceleration due to gravity)
A particle of mass $m$ is projected with velocity $v$ making an angle of ${45^o}$with the horizontal. The magnitude of the angular momentum of the particle about the point of projection when the particle is at its maximum height is (where $g = $ acceleration due to gravity)
Zero
$m{v^3}/(4\sqrt 2 g)$
$m{v^3}/(\sqrt 2 g)$
$m{v^2}/2g$
A particle of mass $m$ is projected with velocity $v$ making an angle of ${45^o}$with the horizontal. The magnitude of the angular momentum of the particle about the point of projection when the particle is at its maximum height is (where $g = $ acceleration due to gravity)
At the highest point, velocity $=v \cos 45^{0}=\frac{v}{\sqrt{2}}$ in horizontal direction
$\text { Momentum }=\frac{m v}{\sqrt{2}}$
$L=$ Angular momentum $=$ Momentum $\times$ Perpendicular distance $L=\frac{m v}{\sqrt{2}(h)}$
Here $h=\frac{v^{2} \sin ^{2} 45^{0}}{2 g}=\frac{v^{2}}{4 g}$
$\therefore L=\frac{m v}{\sqrt{2}} \frac{v^{2}}{4 g}=\frac{m v^{3}}{4 \sqrt{2} g}$
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