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The velocity of the object on earth when it is released from height 'h'.

AV=2gh2V= 2gh^2

BV=2ghV = \sqrt {2gh}

C$$V = \frac{1}{\sqrt {2gh}$$

DV=h22gV = \frac{h^2}{2g}

Answer:

V=2ghV = \sqrt {2gh}

Read Explanation:

The velocity of a freely falling object, under gravity increases at a constant rate. When a body is released from height h, the velocity can be calculated using the equation V=2ghV = \sqrt{2gh}


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