A) \[\sqrt{{{u}^{2}}-2gL}\]
B) \[\sqrt{2gL}\]
C) \[\sqrt{{{u}^{2}}-gl}\]
D) \[\sqrt{2({{u}^{2}}-gL)}\]
Correct Answer: D
Solution :
\[\frac{1}{2}m{{u}^{2}}-\frac{1}{2}m{{v}^{2}}=mgL\] Þ \[v=\sqrt{{{u}^{2}}-2gL}\] \[|\vec{v}-\vec{u}|\,=\sqrt{{{u}^{2}}+{{v}^{2}}}=\sqrt{{{u}^{2}}+{{u}^{2}}-2gL}=\sqrt{2({{u}^{2}}-gL)}\]You need to login to perform this action.
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