A body falls from the top of a building and reaches the ground $2.5 s$ later. How high is the building? (Take $g = 10 m s^{- 2}$ )

30.6 m

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31.25 m

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30 m

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25 m

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Solution

The correct option is B 31.25 m
Height of the building is given by 2nd equation of motion,
$S = u t + \frac{1}{2} g t^{2}$
now since initial velocity is zero so 2nd equation of motion, become,
$S = \frac{1}{2} g t^{2} = \frac{1}{2} \times 10 \times {2.5}^{2} = 31.25 m$

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