A stone is released from the top of a tower of height 19.6 m.

Solution:

Initial velocity of stone, $\mathrm{u}=0 ;$ final velocity of stone,

v = ?

height, $\mathrm{h}=19.6 \mathrm{~m}$; acceleration due to gravity,

$\mathrm{g}=+9.8 \mathrm{~ms}^{-2}$

We know, $v^{2}=u^{2}+2 g h$

$\mathrm{v}^{2}=(0)^{2}+2 \times 9.8 \times 19.6$ or $\mathrm{v}^{2}=19.6 \times 19.6$

$v=\sqrt{19.6 \times 19.6}=19.6 \mathrm{~ms}^{-1}$