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

Here, initial velocity $u=0$

Time $t=2.5\ s$

Gravitational acceleration $g=9.8\ m/s^2$

Let $h$ be the height of the building.

On using the equation of motion, $h=ut+\frac{1}{2}gt^2$

$h=0\times2.5\ s+\frac{1}{2}\times9.8\times(2.5\ s)$

Or $h=0+4.9\times6.25$

Or $h=30.625\ m$

Therefore, the building is $30.625\ m$ high.

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Updated on: 10-Oct-2022

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