A) \[\frac{M}{M+m}\left( \frac{V_{0}^{2}}{2g}-h \right)\]
B) \[{{\left( \frac{M}{M+m} \right)}^{2}}\left( \frac{V_{0}^{2}}{2g}-h \right)\]
C) \[\left( \frac{m}{M+m} \right)\left( \frac{V_{0}^{2}}{2g}-h \right)\]
D) \[{{\left( \frac{m}{M+m} \right)}^{2}}\left( \frac{V_{0}^{2}}{2g}-h \right)\]
Correct Answer: B
Solution :
[b] velocity of acrobat it height his \[V=\sqrt{V_{0}^{2}-2gh}\] Let's say now both (acrobat + monkey) leave the branch velocity is V?. \[MV=(m+M)V'\] \[V'=\left( \frac{M}{m+M} \right)\sqrt{V_{0}^{2}-2gh}\] for maximum height from branch \[\frac{{{(V')}^{2}}}{2g}={{\left( \frac{M}{m+M} \right)}^{2}}\left( \frac{V_{0}^{2}}{2g}-h \right)\]You need to login to perform this action.
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