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Q.

A particle of mass 0.5 kg is projected horizontally with velocity 10 m/s from the top of a tower of height 40 m. The angular momentum of the particle about the base of the tower just after projection is

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a

50  kg−m2/s

b

400  kg−m2/s

c

100  kg−m2/s

d

200  kg−m2/s

answer is D.

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Detailed Solution

(4)

L=mvh=0.5 × 10 × 40  kg−m2/s =200  kg−m2/s

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A particle of mass 0.5 kg is projected horizontally with velocity 10 m/s from the top of a tower of height 40 m. The angular momentum of the particle about the base of the tower just after projection is