Question

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

Answer 1

Given : Height of the tower $H=19.6$ m

Initial velocity of the stone $u=0$

Acceleration $a=g=9.8$ $m{s}^{-2}$

Let the final velocity of the stone be $v$

Using, $v^2-u^2=2aH$

$v^2-0=2\left(9.8\right)\times 19.6$ $⟹v=19.6$ $m{s}^{-1}$