A) 0
B) \[\frac{m{{v}^{3}}}{4\sqrt{2g}}\]
C) \[\frac{m{{v}^{2}}}{\sqrt{2g}}\]
D) \[\frac{\sqrt{2}\,m{{v}^{3}}}{4g}\]
Correct Answer: B
Solution :
Angular momentum J = r x P \[=mv\cos \theta p\] \[=mv\cos \theta \frac{{{v}^{2}}{{\sin }^{2}}\theta }{2g}\] at \[\theta =45{}^\circ \] \[J=\frac{m{{v}^{3}}}{4\sqrt{2g}}\]You need to login to perform this action.
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