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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?

Chetan Mandayam Nayakar
312 Points
10 years ago

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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Chetan Mandayam Nayakar – IIT Madras

509 Points
10 years ago

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