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A particle is projected at time $t=0$ from a point P on the ground with a speed $V_0$ at an angle of $45^{\circ}$ to the horizontal . What is the magnitude of the angular momentum of the particle about point P at time $t= \large\frac{V_0}{g}$

\[\begin {array} {1 1} (a)\;\frac{mV_0^2}{2 \sqrt 2 g} & \quad (b)\;\frac{mV_0^3}{\sqrt 2 g} \\ (c)\;\frac{mV_0^2}{\sqrt 2 g} & \quad  (d)\;\frac{mV_0^3}{2 \sqrt 2 } \end {array}\]
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$ u_x=V_0 \cos \theta=\large\frac{V_0}{\sqrt 2}$
$ u_y=V_0 \sin \theta- gt=\large\frac{V_0}{\sqrt 2}- \frac{V_0}{g}. g$
$\qquad= V_0 \bigg[\large\frac{1- \sqrt 2}{\sqrt 2}\bigg]$
$\overrightarrow{p}=\large\frac{mV_0}{\sqrt 2} \hat i $$+mV_0 \bigg[ \large\frac{1- \sqrt 2}{\sqrt 2}\bigg]\hat {j}$
$\overrightarrow {r} =x \hat i + y\hat j$
$x= u_xt $
$\quad= \large\frac{V_0}{\sqrt 2}. \large\frac{V_0}{g}$
$\quad= \large\frac{V_0^2}{\sqrt 2 g}$
$y= V_0 \sin \theta t -\large\frac{1}{2} $$gt^2$
$\qquad=\large\frac{V_0}{\sqrt 2} \frac{V_0}{g} -\large\frac{1}{2} g \frac{V_o^2}{g^2}$
$\qquad= \large\frac{V_0^2}{g} \bigg[\frac{1}{\sqrt 2}-\frac{1}{2}]$
$\qquad= \large\frac{V_0^2}{g} \bigg[ \large\frac{2 -\sqrt 2}{2 \sqrt 2}\bigg]$
$L= \bar {r} \times \bar {p}$
$\qquad = \bigg\{\large\frac{V_0^2}{\sqrt 2 g} \hat i +\large\frac{V_0^2}{g} \bigg[ \large\frac{2 - \sqrt 2}{2 \sqrt 2}\bigg] \hat j \bigg\}$$ \times \bigg\{\large\frac{mV_0}{\sqrt 2 } \hat i $$+ m V_0 \bigg[\large\frac{1- \sqrt 2}{\sqrt 2}\bigg]$$\hat j \bigg\}$
$\qquad= \large\frac{ mV_0^3}{\sqrt 2 g} \bigg[\large\frac{1- \sqrt 2}{\sqrt 2}\bigg] \hat k - \large\frac{mV_0^3}{\sqrt 2 g} \bigg[ \large\frac{2 - \sqrt 2 }{2 \sqrt 2} \bigg] \hat k$
$\qquad = \bigg(\large\frac{mv_0^3}{2g}-$$ 2 \large\frac{mV_0^3}{2 \sqrt 2 g}\bigg) \hat k - \large\frac{mV_0^3}{ 2 g} \hat k+\frac{mv_0^3}{2 \sqrt 2 g } \hat k $
=> $ \bigg\{ \large\frac{mv_0^3}{2 \sqrt 2 g}-\frac{2 mv_0^3}{2 \sqrt 2}\bigg\} \hat k$
$\qquad= -\large\frac{-mV_0^3}{2 \sqrt 2}$
$|\bar {L}| = \large\frac{mV_0^3}{2 \sqrt 2}$
answered Nov 28, 2013 by meena.p
edited Jun 22, 2014 by lmohan717

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