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

(D) Given: Distance travelled by the stone,$S = 19.6 \, m$.
Initial velocity,$u = 0 \, m/s$.
Acceleration due to gravity,$a = 9.8 \, m/s^2$.
Using the third equation of motion: $v^2 - u^2 = 2aS$.
Substituting the values: $v^2 - 0^2 = 2 \times 9.8 \times 19.6$.
$v^2 = 19.6 \times 19.6$.
$v = \sqrt{19.6^2} = 19.6 \, m/s$.
Therefore,the velocity of the stone just before touching the ground is $19.6 \, m/s$.