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

Answer
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Hint: In this question, we will use the concept of Newton's equation of motion. We will apply the third equation of motion formula for the constant acceleration and substitute the given values to obtain the final velocity of the stone.

Complete step by step solution
The equations of motion are defined as those which are applied to a particle moving linearly, in three dimensions in a straight line with constant acceleration. By constant acceleration we mean that the velocity is changing linearly with time. In mechanics, acceleration is the rate of change of the velocity of an object with respect to time. Accelerations are vector quantities. The orientation of an object's acceleration is given by the orientation of the net force acting on that object.
Now here in this case we have a constant acceleration acting on the body and that is the acceleration due to gravity. $ g = 9.8m/{\sec ^2} $
We will Use the 3rd equation of motion,
 $ \Rightarrow {v^2} = {u^2} + 2as $
Where, $ u $ is the initial velocity of the object
 $ v $ is the final velocity of the object before it just touches the ground
 $ a $ is nothing but the value of $ g $
 $ s $ is the distance covered by the particle from top to bottom of the tower.
Here, we have given that,
 $ \Rightarrow s = 19.6\;{\text{m}} $
 $ \Rightarrow u = 0\;{\text{m/s}} $
Putting these values in the Equation of Motion we get,
 $ \Rightarrow {v^2} = 0 + 2 \times \left( {9.81} \right) \times \left( {19.6} \right) $
We will simplify the above expression as,
 $ \Rightarrow v = \sqrt {384.16} $
After simplification we will get,
 $ \therefore v = 19.6\;{\text{m/s}} $
Therefore, the final velocity of the particle is $ v = 19.6m/\sec $ .

Note
The acceleration was constant that’s why we could use the equation directly. If the acceleration would have been variable, then we would have to integrate that variable function with respect to time and the usage of a different equation of motion would have been there. The other equations of motion are-
 $ \Rightarrow v = u + at $
 $ \Rightarrow s = ut + \frac{1}{2}a{t^2} $ .