A) \[R/2\]
B) \[R/3\]
C) \[R/5\]
D) \[R/9\]
Correct Answer: D
Solution :
Key Idea: Escape velocity\[{{v}_{e}}=\sqrt{2gR}\] Let \[h\] be the maximum height attained, then from equation of motion \[{{v}^{2}}={{u}^{2}}+2gh\] When, \[u=0,\,\,v=\sqrt{2gh}\] Given, \[v=\frac{{{v}_{e}}}{3}\] where, \[{{v}_{e}}=\sqrt{2gR}\] \[\therefore \] \[\sqrt{2gh}=\frac{1}{3}\sqrt{2gR}\] Squaring both sides of equation, we get \[h=\frac{R}{9}\]You need to login to perform this action.
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