Question

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

Answer 1

Initial velocity $u=0$ (As body dropped from rest)

And final velocity $=20m{s}^{-1}$

$a=g=10m{s}^{-2}$

Let $h$ be the height of the tower.

As we know that third equation of motion is,

$v^2-u^2=2as$

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

$400=20h$

$h=20m$

Hence, the height of the tower is equal to $20m$.