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a particle of mass m is projected with a velocity v making an angle of 45 degrees with the horizontal. the magnitude of the angular momentum of the projectile about the point pf projection when the particle is at its max height
Dear Pranay
max height ,H= v2sin2θ/2g
velocity at max height = ux= vcosθ
angular momentum= muxH
and putting θ =450
angular momentum = mv3/g4√2
All the best.
AKASH GOYAL
AskiitiansExpert-IITD
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if the point of projection is taken as origen then
coordinates of maximum height (H,R/2) or r =Hi + (R/2)j in vector form
R is horizontal range and H is maximum height ....
at maximum height there is only horizontal component of velocity no vertical component exists ...
so , V= vcos@i ......
linear momentam = mvcos@i ( @ is the angle of projection)
we know that angular momentam(L) about point of projection = r*p (* is cross product)
L =(Hi + (R/2)j)*(vi)
L =Rv/2 (-K)
L=v3 /2g (-k) ( R =v2 /g at @=45 degree)
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