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A particle of mass m is projected from the ground with an initial speed u at an angle α, The magnitude of its angular momentum at the highest point of its trajectory about the point of projection is mu3cosαsin2αxg. Find x.

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Solution

Refer figure, vertical distance, of particle from point of origin, when at highest point of its trajectory is H=(usinα)2/2g.
The velocity at highest point is vh=ucosα.
So the angular momentum is mvh×H=mu3cosαsin2α/2g
so, x=2

401715_215949_ans_a6e88db417ba4b058889a8873982d1d3.png

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