Particle of mass m is projected with a velocity v 0 making an angle of 45 0 with horizontal. The magnitude of angular momentum of the projectile about the point of projection at its maximum height is
A
Zero
B
C
D
Text Solution
Verified by ExpertsThe correct answer is:
D
Speed of the particle at the top = horizontal component of the speed of projection
⇒ ⇒ v = v 0 cos θ θ 0 = v 0 cos 45 0 = 
The angular momentum the particle about O = L = mvr sin θ θ where θ θ = angle between 
⇒ ⇒ L = mv h ( 
r sin θ θ = h)
Putting h = 
we obtain
L = 
,
Putting v = 
We obtain L = 
. Hence
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