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pranay -askiitians expert Grade: 12
        

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 

6 years ago

Answers : (2)

AKASH GOYAL AskiitiansExpert-IITD
419 Points
										

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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6 years ago
vikas askiitian expert
510 Points
										

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)

6 years ago
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