Question: Suppose a question states:

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> A ball is dropped from a height of 2m. Find the time it takes for which the ball is in free fall.
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To solve this, we would plug in our known values into the kinematic equations.

But if the distance between the ball and the ground is 2m, then as soon as the ball travels 2m, at that exact moment it will touch the ground. So it will experience some deceleration at the 2m mark and thus, its acceleration will not be constant.

But to use kinematic equations, we must have constant acceleration motion. So why is using 2m as our displacement in the kinematic equations still correct? Shouldn't we instead use a distance that is JUST above the ground?

Answer: 
You're correct that the distance fallen should be $x = 2 \text{ m} - \epsilon$, where $\epsilon$ is some infinitesimally small distance. You could then solve the problem for the free-fall time as a function of $\epsilon$; you could call this function $t\_\text{fall}(\epsilon)$.

But if you did this, you'd find that as $\epsilon \to 0$, $t\_\text{fall}$ approaches the value that you'd get for the ball to freely fall 2 meters, without worrying about what happens *at* $x = 2 \text{m}$. In other words, the time required to fall "to a point just above the ground" is effectively the same as the amount of time required to fall "to the ground", and so we usually treat these two things as the same.
Here, if you consider the ball with a certain diameter then take a respective point on the ball to be easy to understand

Let's take bottom most point of the ball,ball is at the point of release still not released.

Now the distance or straight line or displacement if you consider a straight free fall it is as said 2m vertical free fall.

Now the time needs to be calculated for the fall meaning the point we considered must be touching the top most like literally we consider the top most point on the ground neglecting any rough surfaces .

Then from point of release to UNTIL THE BOTTOM MOST POINT OF THE BALL TOUCHES THE GROUND'S TOP MOST SURFACE POINT BALL IS UNDER ONLY INFLUENCE OF GRAVITATION OF EARTH NEGLECTING OTHER DRAG FORCES LIKE AIR AND ETC ETC.

SO THATS WHY SINCE BALL IS UNDER CONSTANT UNIFORM ACCELERATION WE WILL USE THE EQUATIONS OF UNIFORMLY ACCELERATED BODY.