Question

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

Answer 1

Hint
We use the third equation of motion to solve this question. It is also known as the law of constant acceleration. The third equation of motion gives the final velocity of an object under uniform acceleration given the distance travelled and an initial velocity. We take the gravity of earth as the acceleration because the stone is being dropped from a height on earth. Initial velocity is taken as zero because the stone is being released.
$\Rightarrow {v^2} - {u^2} = 2as$
Where final velocity is represented by $v$, Initial velocity is represented by $u$, Constant acceleration is represented by$a = g$, acceleration is due to gravity, and Distance travelled is represented by $s$.

Complete step by step answer
We are given the question that the height of the tower is $19.6m$, which is the distance travelled by the stone $(s)$.
We take initial velocity $(u)$ as zero because the stone is being released which means the stone was at rest.
The stone is dropped, so the acceleration is caused due to gravity $(g)$ which is $9.8m/s$.
Now the only unknown in the equation is the final velocity ($v$).
Substituting all the values in the equation, ${v^2} - {u^2} = 2as$, we get,
${v^2} - {0^2} = 2 \times g \times 19.6 $
$\Rightarrow{v^2} = 2 \times 9.8 \times 19.6 $
$\Rightarrow v = \sqrt {384.16} $
$\Rightarrow v = 19.6m/s $
Hence the final velocity of the stone just before touching the ground is $v = 19.6m/s$

Note
In such questions, the student might read the question and wonder what the initial velocity is. If the question mentions any statement about the object starting from rest or being released/dropped then the initial velocity is taken as zero. If the object is being accelerated by no other means but gravity, then we take acceleration as gravity.