What is the formula to calculate the distance covered in a specific time (e.g 3rd second, 4th sec...) by a contantly accelarated body? consider the following question:
The distance covered by a freely falling body in 2nd second of its journey is?
solution:
distance covered is 1st second is:
S1=Vit + 1/2gt^2...put values
S=5m
Distance covered in 2nd second is:
S2=Vit + 1/2gt^2...put Vi=10m/s^2...(this value is calculated from 1st equation of motion) by putting all values we get, 
S=15m...hence (S2=3S1)
i would like a single equation which solves this stuff! so how to derive a single equation instead of using two equation of motions?
 A: I hope that the following answer helps.
Let the distance covered in $n$ seconds be $S_n$ and distance covered in $n-1$ seconds be $S_{n-1}$.
Then considering $1D$ motion,
$S_n = u_0n + \frac{an^2}{2}$
$S_{n-1} = u_0(n-1) + \frac{a(n-1)^2}{2}$
$S_{n} - S_{n-1} = u_0(n) + \frac{an^2}{2} - u_0(n-1) - \frac{a(n-1)^2}{2}$
Hence, 
$$S_{n^{th} second} = u_0 + \frac{a(2n-1)}{2}$$
where $a$ is the acceleration and $u_0$ is the initial velocity.
A: Your question is somewhat unclear, so I will answer what I understood. 
You wish to find the position at any time of a freely falling object.
In order to do that, you must find the position as a function of time. Now, the motion will depend on the following variables. The initial position or height, h, the initial velocity, $v_0$ and the acceleration a. Thus the general formula is 
$$y(t) = h +v_0t +\frac{at^2}{2}$$
Note that in free fall the acceleration a is just the acceleration of gravity g=-9.8m/s^2
This should be the general formula that you are looking for. I recommend you to check out books such as Serway physics for scientists an engineers if you wish to know more
