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The ratio of the magnitudes of the angular momentum of a projectile, at the highest point on trajectory and that at the point where it strikes the ground, about the point of projection is?
Dear Varsha,
m((u2sin2(theta)/2g)*ucos(theta))/m((2u2sin(theta)cos(theta)/g)*usin(theta)=(1/2)/2=1/4
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Askiitians Expert
Chetan Mandayam Nayakar – IIT Madras
angular momentam = mr*v
at highest point V = vcos@(i) & r = R/2(i) + H(j)
R is range & H is max height of projectile..
angular momentam=m(r*v)=mvcos@H (-k) ...............1
when the particle strikes the ground then its velocity v = vcos@(i)+vsin@(-j)
r = R(i) + 0(j)
angular momentam=m(r*v)
=mvRsin@(-k)............2
ratio = eq1/eq2=cos@H/sin@R=1/4
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