A particle of mass m is projected at t = 0 from a point O on the ground with a speed vq at an angle of 45° to the horizontal. The magnitude of the angular momentum of the particle about O at the time t = v 0 /g is close to [Take 
= 1.4]
Text Solution
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Using 1 st-equation of motion v = u + at along horizontal direction,
[as a=0]
and along vertical direction
So, the velocity of particle at time t = (vo/g) in vector form will be
....(i)
Now from 2nd equation of motion 
along horizontal direction,
and along vertical direction,
So the position vector 
at the t (= v 0 /g) will be
....(ii)
Now as by definition 
So 
i.e 
i.e., angular momentum has magnitude 
and is directed along negative z-axis.
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