A stone falls from a building and reaches the ground 2.5 seconds later. How high is the building ? $(g = 9.8 m / s^{2})$

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Solution

Given : Initial velocity u = 0 m/s

gravity acceleration g = 9.8 m/s^2

height of building h = ?

time t = 2.5 sec

from 2nd equation of motion

$h = u t + \frac{1}{2} g t^{2}$

$h = 0 \times 25 + \frac{1}{2} \times 9.8 \times 2.5 \times 2.5$
$h = \frac{1}{2} \times 9.8 \times 2.5 \times 2.5$
$h = \frac{61.25}{2} = 30.625 m$ .

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