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Question

Physics Question on Motion in a straight line

A particle is released from a height equal to radius of Earth. Find its velocity when it strikes the ground.

A

gR\sqrt{gR}

B

gR2\frac{\sqrt{gR}}{2}

C

2gR\sqrt{2gR}

D

4gR\sqrt{4gR}

Answer

gR\sqrt{gR}

Explanation

Solution

Hence the correct option is (A). gR\sqrt{gR}

From mechanical energy conservation,

−GMm2R+0=−GMmR+12mv2-\frac{GMm}{2R}+0=-\frac{GMm}{R}+\frac{1}{2}mv^2

⇒v2=GMR=gR⇒gR\Rightarrow v^2=\frac{GM}{R}=gR\Rightarrow \sqrt{gR}