Question

# A stone is released from the top of a tower of height $19.6m$. Calculate its final velocity just before touching the ground.

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Solution

## Step 1: Given:Height of the tower = $19.6m$Acceleration due to gravity $g=9.8m{s}^{-2}$Step 2: Formula and calculation:Using the third equation of motion${v}^{2}={u}^{2}+2gs\phantom{\rule{0ex}{0ex}}{v}^{2}=0+2×9.8×19.6\phantom{\rule{0ex}{0ex}}{v}^{2}=19.6×19.6\phantom{\rule{0ex}{0ex}}v=19.6m{s}^{-1}$Hence, the speed of stone just before touching the ground is $19.6m{s}^{-1}$

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