A) zero
B) \[\frac{3m{{v}^{3}}}{16g}\]
C) \[\frac{\sqrt{3}m{{v}^{3}}}{16g}\]
D) \[\frac{3m{{v}^{3}}}{8g}\]
Correct Answer: B
Solution :
Maximum height, \[H=\frac{{{v}^{2}}{{\sin }^{2}}{{60}^{o}}}{2g}\] \[=\frac{{{v}^{2}}}{2g}\times \frac{3}{4}=\frac{3{{v}^{2}}}{8g}\] Momentum of particle at highest point \[p=mv\cos \,{{60}^{o}}=\frac{mv}{2}\] Angular momentum = pH \[=\frac{mv}{2}\times \frac{3{{v}^{2}}}{8g}\] \[=\frac{3m{{v}^{3}}}{16g}\]
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