A) \[200 m/s\]
B) \[150 m/s\]
C) \[400 m/s\]
D) \[300 m/s\]
Correct Answer: A
Solution :
| Initial, \[K.E.=\frac{1}{2}m{{v}^{2}}=\frac{1}{2}\times \frac{20}{1000}\times 600\times 600=3600\,J\] |
| Change in \[K.E.=P.E.\] |
| \[\frac{1}{2}m({{v}^{2}}-v{{'}^{2}})=mgh\] |
| \[\Rightarrow \,\,3600-\frac{1}{2}\times \frac{20}{1000}\times {{v}_{1}}^{2}=4\times 10\times 80\] |
| \[\Rightarrow \,\,\,\,{{v}_{1}}=200m/s\] |
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