Question

A free falling body travels_ _ _ of total distance in ${5^{th}}$ second.
A. $8\%$
B. $12\%$
C. $25\%$
D. $36\%$

Answer 1

Calculate the total percentage of the distance travelled by the freely falling body by using two cases, that is by considering the body travelling for $5$ seconds and the body travelling for $4$ seconds and then subtract them to find the distance of the ${5^{th}}$ second.

Formula used:
Laws of motion:
$S = ut + \dfrac{1}{2}a{t^2}$

Complete step by step answer:
We know that a free falling body will travel with a certain percentage of total distance in a given time. Therefore, some data is given in the question for calculating the exact percentage of total distance travelled.
Let us assume that the free falling body is travelling a distance ${S_2}$in 5 seconds.
Then by using the laws of motion, we can write
$\therefore S = ut + \dfrac{1}{2}a{t^2}$ . . . (1)
Where, $u$is the initial velocity.
$t$ total time taken
$a$ is acceleration of the body
Now, for a freely falling body,
Initial velocity, $u = 0$
And the acceleration will be replaced by the acceleration due to gravity i.e. $a = - g$
So, equation (1) will become
$- {S_2} = - \dfrac{1}{2}g{t^2}$ (since, the direction of the body is downwards)
$\Rightarrow {S_2} = \dfrac{1}{2}g{t^2}$
So, for 5 seconds, we get
${S_2} = \dfrac{1}{2} \times 10 \times {5^2}$ (since $g = 10m/{s^2}$)
$\Rightarrow {S_2} = 5 \times 25$
$\Rightarrow {S_2} = 125m$ . . . (2)
Now, let us assume that the distance travelled after 4 seconds is ${S_1}$
Then, for 4 seconds, we get
${S_1} = \dfrac{1}{2} \times 10 \times {4^2}$
$\Rightarrow {S_1} = 5 \times 16$
$\Rightarrow {S_1} = 80m$ . . . (3)
Therefore, the distance travelled in the ${5^{th}}$second will be
${S_2} - {S_1} = 125 - 80 = 45m.$
Therefore, the total percentage distance travelled by the freely falling body in the${5^{th}}$seconds will be $\dfrac{{{S_2} - {S_1}}}{{{S_2}}} \times 100$
$= \dfrac{{45}}{{125}} \times 100 = 36\% .$
Hence the correct option is (D) $36\%$

Note: Since, the speed of the freely falling body will keep on increasing with time, we cannot just calculate the distance travelled by the body in the ${1^{st}}$ second and assume that it will be the same for every passing second.