**Question : **

Two bodies begin a free fall from the same height at a time interval of N s. If vertical separation between the two bodies is 1 m after n seconds from the start of the first body, then what is n equal to?

**Solution** :

Hint: Assume the necessary variables and find the relations based on the data given in the problem statement. Using the second equation of motion, find the answer.

Complete step-by-step answer:

Let the height from which the two bodies are thrown be = $h$ m

Let us say that the first body fell at time $t_1$ and the second body fell at time $t_2$.

Now, it is given that second body fell after an interval of N s compared to the first body.

$\therefore t_2 – t_1 = N \to (1)$

After $n$ s, the time passed for the first body in free fall would be $= n – t_1$

After $n$ s, the time passed for the second body in free fall would be $= n – t_2$

But from eqn (1), we can write $t_2 = N + t_1$

$\Rightarrow n – t_2 = n – (N + t_1) = n – N – t_1$

For simplicity in calculation, let us assume that, the first body was thrown at 0 s i.e. $t_1$ = 0.

We will get,

The time passed for the first body in free fall $= n – 0 = n$

The time passed for the second body in free fall $= n – N – 0 = n – N$

We know the second equation of motion, i.e.

$s = u t + \dfrac{1}{2} a t^2$, where s is the distance traveled, t is the time, a is acceleration and u is the initial velocity.

It is told bodies begin free fall, so that means the initial velocity $u$ for both bodies would be 0.

The acceleration is the acceleration due to gravity i.e. $a = g$

Now, distance travelled by first body = $s_1 = \dfrac{1}{2} g n^2$

Distance travelled by second body = $s_2 = \dfrac{1}{2} g (n-N)^2$

It is given that after $n$ s, the distance between them is 1 m.

$s_1 – s_2 = 1 = \dfrac{1}{2} g n^2 – \dfrac{1}{2} g (n-N)^2$

$\Rightarrow \dfrac{1}{2} g [n^2 – (n – N)^2] = 1$

We know that $(a-b)^2 = a^2 + b^2 – 2ab$, we get,

$\Rightarrow n^2 – n^2 – N^2 + 2nN = \dfrac{2}{g}$

$\Rightarrow 2nN – N^2 = \dfrac{2}{g}$

$\Rightarrow 2nN = N^2 + \dfrac{2}{g}$

$\Rightarrow n = \dfrac{N}{2} + \dfrac{1}{gN}$

It is the required value of n.