Question

a particle of mass 1 kg is projected from ground with velocity 20ms1 at 45from horizontal. magnitude of its angular momentum about point of projection when it is at maximum height. (g=10ms2).

Answer 1

Angular momentum = m(r x v)

It's defined as the cross product of position vector and momentum.

Now, in a projectile we know that at maximum height point , the velocity is only acting in the x-direction, whose magntiude will be v cos theta which in this case will be

20(1/√2) = 10√2 m/s

Now,

Because, for now we want the magntiude

Thus, Angular momentum = r(mv) sin theta

So, r sin theta on geometric analysis will come out to be the maximum height.

Max height = v^2 sin^2 theta/2g for a projectile

= (400)(1/2)/20 = 10 metres

Now,

Angular momentum = mv(H) = (1)(10√2)(10) = 100√2 in magntiude

And by applying the Right hand thumb rule for finding the direction of Angular momentum the direction will come out to be radially inwards of the plane if the projectile is considered to be in plane of paper sheet.

Answer 2

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