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varsha rao Grade: 11
        

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?

6 years ago

Answers : (2)

Chetan Mandayam Nayakar
312 Points
										

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

6 years ago
vikas askiitian expert
510 Points
										

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

6 years ago
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