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

Given, u = 0 m/s

\( h = s = 19.6 m \)

g = 9.8 m/s2 (falling down)

We know that, \( v^2 – u^2 = 2 g s \)

\( \Rightarrow v^2 – (0)^2 = 2 x 9.8 x 19.6 \)

\(\Rightarrow \) v = 19.6 m/s

The final velocity just before touching the ground is 19.6 m/s.