Question

What will be the ratio of the distance moved by a freely falling body from rest in 4th and 5th seconds of journey?
A. $4:5$
B. $7:9$
C. $16:25$
D. $1:1$

Answer 1

Hint:In order to answer this question, first we will write the formula of distance covered in $nth$ second in terms of velocity and acceleration. And then we will put $n = 4$ and $n = 5$ separately to find the value of distance moved by a freely falling body from rest in 4th and 5th seconds of journey. Now, we can easily find their ratio.

Complete step by step answer:
First we will write the equation of the distance covered in $nth$ second is given by:
We will apply the formula of distance covered in terms of velocity and acceleration:-
${s_n} = u + \dfrac{a}{2}(2n - 1)$ ……….eq(I)
The body is moving freely falling from the rest, so the velocity is:
$u = 0\,m.{s^{ - 1}}$
And the body is falling, then the acceleration is the gravity itself:
$a = g$
Now, if in 4th second of journey, $n = 4$ in equation(i):
${S_4} = \dfrac{g}{2}(2 \times 4 - 1) = \dfrac{{7g}}{2}$

Again, if in 5th second of journey, $n = 5$ in equation(i):
${S_5} = \dfrac{g}{2}(2 \times 5 - 1) = \dfrac{{9g}}{2}$
Now, the ratio of the distance moved by a freely falling body from rest in 4th and 5th seconds of journey:-
$\dfrac{{{S_4}}}{{{S_5}}} = \dfrac{{\dfrac{{7g}}{2}}}{{\dfrac{{9g}}{2}}} \\
\Rightarrow \dfrac{{{S_4}}}{{{S_5}}} = \dfrac{7}{9} \\ $
$\therefore {S_4}:{S_5} = 7:9$ .
Therefore, $7:9$ is the ratio of the distance moved by a freely falling body from rest in 4th and 5th seconds of journey.

Hence, the correct option is B.

Note:A body that is falling downhill only due to gravitational force is referred to as a freely falling body. The initial velocity of a freely falling body is zero, and the body moves with an acceleration equal to the positive acceleration due to gravity.