A ball is dropped downwards, after $1\sec $ another ball is dropped downwards from the same point. What is the distance between them after $3\sec $ ? (Take $g = 10m{s^{ - 2}}$)

Hint: We will solve the given question using the second equation of motion, which is also known as position-time relation. Using that first we will find the distance travelled by the balls individually, and then simply subtract the obtained values to find the required answer.

Formula used:
$s = ut + \dfrac{1}{2}a{t^2}$

Complete step-by-step answer:
Distance at a given time is given by the second equation of motion,
For first body, ${u_1} = 0$ , $s = {s_1}$ , $a = g$ , ${t_1} = 3\sec $
So, ${s_1} = \dfrac{1}{2}g \times 9$
Now, for second body, ${u_2} = 0$ , $s = {s_2}$ , $a = g$, ${t_2} = 2\sec $
So, ${s_2} = \dfrac{1}{2}g \times 4$
Therefore, the difference between the two body’s after $3\sec $ is given by,
${s_2} - {s_1} = \dfrac{1}{2}g \times 5$
If $g = 10m{s^{ - 2}}$
then ${s_2} - {s_1} = \dfrac{1}{2}(10) \times 5$ $ = 25m$
Hence, the distance between the balls after $3\sec $ is $25m$.

Additional Information: In the given question, we have used the second equation of motion. Using this formula we can calculate the displacement when the values for time, acceleration and initial velocity are known.
In case of free fall, this equation can be written as $s = ut + \dfrac{1}{2}g{t^2}$
Where, $s$ is the height from which the object falls on the ground, $g$ is the acceleration due to gravity. Value of $g$ is taken negative when the object is thrown upward.

Note: Free fall is defined as the motion of a body where its weight is the only force acting on the object. The acceleration of the free-falling objects is known as the acceleration due to gravity, since objects are pulled towards the centre of the earth. The acceleration due to gravity is found to be constant on the surface of the Earth and has the value of $9.8m/{s^2}$. The value of $g$ is sometimes taken as $10$ to help minimize the calculations and get an approximate answer.