Question #a7296

1 Answer
Feb 7, 2017

#(mv^3sqrt2)/(8 g)#

Explanation:

We know that angular momentum #vecL# of a particle of mass #m# moving with velocity #vecv# with respect to a selected origin is given by the equation

#vecL=vecrxxvecp#
where #vecp# is linear momentum and #=mvecv#
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The velocity of the particle changes continuously due to acceleration due to gravity #g# acting on the vertical component of velocity which #=vsin45^@=v/sqrt2#.
Similarly horizontal component of velocity#=vcos45^@=v/sqrt2#

We know that at maximum height vertical component of velocity is zero.
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(Please read initial velocity as #v# instead of #"u"# in the figure, #theta=45^@#)

Using the kinematic equation
#v^2-u^2=2as# and inserting given values we get
#0^2-(v/sqrt2)^2=2(-g)h#
#=>h=v^2/2xx1/(2g)#
#h=v^2/(4g)#

Now #|vecL|=hxxm|vecvcostheta|#
Inserting calculated values we get

#|vecL|=v^2/(4g)xxmxxv/sqrt2#
#=>|vecL|=(mv^3sqrt2)/(8 g)#