A particle of mass m is projected with a velocity V making an angle of 45^o with horizontal. magnitu

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3-2.Motion in Plane

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A particle of mass $m$ is projected with a velocity $V$ making an angle of $45^o$ with the horizontal. The magnitude of the angular momentum of the projectile about the point of projection when the particle is at its maximum height $h$ is

A

zero

B

$\frac{{m{V^2}}}{{\sqrt 2 \,g}}$

C

$\frac{{m{V^2}}}{{4\sqrt 2 \,g}}$

D

$m\sqrt {2g{h^3}} $

Solution

Momentum at the highest point

$=m V \cos 45^{\circ}=\frac{m V}{\sqrt{2}}$

Maximum height $=h$

$h=\frac{V^{2} \sin ^{2} 45^{\circ}}{2 g}=\frac{V^{2}}{4 g}$

Angular momentum $=L$

$L=$momentum$\times$height$=\frac{m V}{\sqrt{2}} \times \frac{V^{2}}{4 g}=\frac{m V^{3}}{4 \sqrt{2} g}$

But $\quad \frac{V^{2}}{4 g}=h \quad$ or $\quad V=\sqrt{4 g h}$

$L=\frac{m}{4 \sqrt{2}} \frac{(4 g h)^{3 / 2}}{g}=m \sqrt{2 g h^{3}}$

Standard 11

Physics

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