A) zero
B) \[\frac{m{{\upsilon }^{3}}}{\sqrt{2}g}\]
C) \[\frac{\sqrt{3}}{16}\frac{m{{\upsilon }^{3}}}{g}\]
D) \[\frac{\sqrt{3}}{2}\frac{m{{\upsilon }^{2}}}{g}\]
Correct Answer: C
Solution :
\[{{L}_{0}}={{\Pr }_{\bot }}\] \[{{L}_{0}}=mv\cos \theta H\] \[=mg\frac{\sqrt{3}}{2}.\frac{{{v}^{2}}{{\sin }^{2}}{{30}^{o}}}{2g}=\frac{\sqrt{3}m{{v}^{3}}}{16g}.\]You need to login to perform this action.
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