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A particle is projected at time $t=0$ from a point P on the ground with speed u at an angle $\theta$ with the horizontal. The angular momentum of the particle about the origin when it is at the maximum height

\[\begin {array} {1 1} (a)\;u \cos \theta . \frac{u^2 \sin 2 \theta}{g} & \quad (b)\;\frac{u^3 \cos \theta. \sin 2 \theta }{2g} \\ (c)\;\frac{u^3 \cos \theta. \cos 2 \theta }{g} & \quad  (d)\;\frac{u^3 \sin 3 \theta}{2g} \end {array}\]

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1 Answer

$\overrightarrow{L} =\bar {r} \times \bar {p}$
$| \bar {L}|= mu \cos \theta \times H_{max}$
$\qquad= \large\frac{mu \cos \theta. u^2 \sin 2 \theta}{g}$
answered Nov 28, 2013 by meena.p
edited Jun 21, 2014 by lmohan717

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