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The escape speed of an object on the surface of the earth is V. If the object is thrown out with speed 4V from the surface of the earth, the speed of the object far away from the earth is

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3 V

$\sqrt{15} V$

2.5 V

$\sqrt{8} V$

answer is B.

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Detailed Solution

$V_{\infty} = \sqrt{v^{2} - {v_{e}}^{2}}$
$\sqrt{16 v^{2} - v^{2}} = \sqrt{15} v$

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