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A body dropped from height 'h' reaches the ground with velocity 'v'. If dropped from '4h', it reaches with:

A2v
B4v
Cv/2
Dv^2

✓ Correct Answer: A2v

v = sqrt(2gh). If h becomes 4h, v becomes sqrt(2g*4h) = 2*sqrt(2gh) = 2v.

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