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A ball of mass 160g is thrown up at an angle of 60° to the horizontal at a speed of 10 ms-1. The angular momentum of the ball at the highest point of the trajectory with respect to the point from which the ball is thrown is nearly g=10 ms-2

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a

1.73 kg m2/s

b

3.0 kg m2/s

c

3.46 kg m2/s

d

6.0 kg m2/s

answer is B.

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Detailed Solution

Angular momentum L=r→×mv→=Hmvcosθ=v2sin2θ2 gmvcosθ [H=v2sin2θ2 g]=102×sin260°×0.160×10×cos60°2×10=3 kg m2/s

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