A) \[\frac{\sqrt{3}}{2}\frac{m{{v}^{2}}}{g}\]
B) zero
C) \[\frac{m{{v}^{3}}}{\sqrt{2}g}\]
D) \[\frac{\sqrt{3}}{16}\frac{m{{v}^{3}}}{g}\]
Correct Answer: D
Solution :
Angular momentum of the projectile \[L=m{{v}_{h}}r\bot =m(v\cos \theta )h\] (where h is the maximum height) \[p=m(v\cos \theta )\left( \frac{{{v}^{2}}\,{{\sin }^{2}}\theta }{2g} \right)\] \[L=\frac{m{{v}^{3}}\,{{\sin }^{2}}\theta }{2g}=\frac{\sqrt{3}}{16}\,\,\frac{m{{v}^{3}}}{g}\]
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