
The angular momentum L=r×p
At maximum height angular momentum about O,
L=mvcosθH
=mvcosθ×2gu2sin2θ(forhorizontalv=u)
=2gmu3sin2(60∘)cos(60∘)
=3kgm2s−1
A ball of mass 160g is thrown up at an angle of 60∘ to the horizontal at a speed of 10ms−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=10ms−2)
Held on 19 Apr 2014 · Verified 6 Jul 2026.
1.73kgm2s−1
3.46kgm2s−1
3.0kgm2s−1
6.0kgm2s−1
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