A) zero
B) \[\frac{mv{{h}^{2}}}{\sqrt{2}}\]
C) \[\frac{m{{v}^{2}}h}{2}\]
D) \[\frac{mv{{h}^{3}}}{\sqrt{2}}\]
E) \[\frac{mvh}{\sqrt{2}}\]
Correct Answer: E
Solution :
When a particle is projected with a speed v at \[45{}^\circ \] with the horizontal then velocity of the projectile at maximum height. \[v=v\cos 45{}^\circ =\frac{v}{\sqrt{2}}\] Angular momentum of the projectile about the point of projection \[=mvh\] \[=m\frac{v}{\sqrt{2}}h=\frac{mvh}{\sqrt{2}}\]
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