Q,  Aparticle is projected at time t = 0 from a point P with a speed V at an angle 45 to horizontal. Find the magniyude and direction of angular momentum of the particle about the point P at tome t = V / g.

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plz solve.

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FOR HORIZONTAL MOTION

VX=VCOS 45=V/ROOT 2

X=VXT=V/ROOT 2*V/G=V^2/2g

FOR VERTICAL MOTION

Vy=VSIN 45-1/2AT^2=V/ROOT 2-V=1-ROOT2/ROOT 2 *V

Y=VyT-1/2AT^2=V^2/2g*(ROOT 2-1)

THE ANGULAR MOMENTUM

IS MVR

M(iX+JY)*(IVx+JVy)

=M(KXVy-KYVx)

=-KMV^3/2ROOT 2*2 G

THUS THE ANGULAR MOMENTUM IS IN THE NEGATIVE Z AXIS

AND ITS VALUE IS MV^3/2ROOT 2*2 G

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FOR VERTICAL MOTION

Vy=VSIN 45-1/2AT^2=V/ROOT 2-V=1-ROOT2/ROOT 2 *V

think it's typo error.