A particle of mass m is projected at 45^o at | vedclass

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Question

$	au=\operatorname{mg} \mathrm{x}=$ torque

$	au=\operatorname{mg}\left(V_{0} \cos 45ight) t$

$\frac{d J}{d t}=\frac{m g V_{0}}{\sqrt{2}} t$

Vedclass · Accepted Answer

$\frac{1}{2 \sqrt 2} \frac{mV_0^3}{g}$

Vedclass · Answer

$	au=\operatorname{mg} \mathrm{x}=$ torque

$	au=\operatorname{mg}\left(V_{0} \cos 45ight) t$

$\frac{d J}{d t}=\frac{m g V_{0}}{\sqrt{2}} t$

$\int_{0}^{\mathrm{J}} \mathrm{d} \mathrm{J}=\int_{0}^{\mathrm{t}} \frac{\mathrm{mg} \mathrm{V}_{0}}{\sqrt{2}} \mathrm{t} \mathrm{dt}$

$J=\frac{m g V_{0}}{\sqrt{2}} \frac{t^{2}}{2}$

$J=\frac{m g V_{0}^{3}}{2 \sqrt{2} g^{2}}$

$J=\frac{1}{2 \sqrt{2}} \frac{m V_{0}^{3}}{g}$

A particle of mass $m$ is projected at $45^o$ at $V_0$ speed from point $P$ at $t = 0$. The angular momnetum of particle about $P$ at $t = \frac{V_0}{g}$ is:-

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