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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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