Question

A body is dropped from the top of a tower It acquires a velocity $20m{s}^{-1}$ on reaching the ground. Calculate the height of the tower (Take $g=10m{s}^{-2}$)

Answer 1

Given,

$u=0m/s$

$v=20m/s$

$g=10m/s^2$

From 3rd equation of motion,

$2gh=v^2-u^2$

$h=\frac{v^2-u^2}{2g}$

$h=\frac{20\times 20-0\times 0}{2\times 10}$

$h=20m$

The height of the tower is $20m$.