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Find the average velocity of a particle released from rest from a height of 125 m over a time interval till it strikes the ground. Take $g = 10 m / s^{2}$ .

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$25 m / s$

$2 m / s$

$5 m / s$

$10 m / s$

answer is D.

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Detailed Solution

$t = \sqrt{\frac{2 h}{g}} = \sqrt{\frac{2 \times 125}{10}} = 5 s$

$v_{α v} = \frac{s}{t} = \frac{125}{5} = 25 m / s$

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