A ball of small size is released from height $R_{E}$ , find the speed of ball when it reaches the surface of earth. Assume $R_{E}$ is the radius of earth and $g$ is acceleration due to gravity on earth's surface.

\sqrt{2 g R_{E}}

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\sqrt{g R_{E}}

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\sqrt{\frac{g R_{E}}{2}}

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None

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Solution

The correct option is B $\sqrt{g R_{E}}$
$K . E . + P . E . = c o n s t a n t$
$0 - \frac{G M m}{2 R_{E}} = \frac{1}{2} m v^{2} - \frac{G M m}{R_{E}}$
$\frac{G M m/}{R_{E}} - \frac{G M m/}{2 R_{E}} = \frac{1}{2} m/ v^{2}$
$\frac{G M}{/ 2 R_{E}} = \frac{v^{2}}{/ 2}$
$v^{2} = \frac{G M}{R_{E}} \frac{R_{E}}{R_{E}} \Rightarrow v = \sqrt{g R_{E}}$

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