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?

Answer Verified Verified

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.


Note: To solve such problems, one should try to visualise the problem and proceed systematically. One should choose the right equations to use. Mistakes can be avoided while solving and forming equations. Data extraction from the question statement is very important to reach the answer.

Bookmark added to your notes.
View Notes