A body falls freely from rest. It covers as much distance in the last second of its motion as covered in the first three seconds. the body has fallen for time of:
A) $3\,s$
B) $5\,s$
C) $7\,s$
D) $9\,s$

Hint: Here, we know the meaning of the distance traveled. Use a kinematic equation to fix this problem, then. You should use the free fall body idea that the free-fall body is just a body under the control of gravity. Do not recognize the air temperature resistance.

Formula used:
Distance covered by body in seconds
$D = \dfrac{{g \times {t^2}}}{2}$
Where,
$D$ is distance,$g$ is force of gravity,$t$ is time

Complete step by step solution:
Given by,
A body falls freely from rest,
Distance covered by body in three seconds,
${D_1} = \dfrac{{g \times {3^2}}}{2}$
Distance covered by body in last seconds,
${D_2} = {\dfrac{{g \times t}}{2}^2}$
Distance covered by body one before influence,
${D_3} = \dfrac{{g \times {{\left( {t - 1} \right)}^2}}}{2}$
According to that ,
A given equation is written as,
$\Rightarrow$ ${D_1} = {D_3} - {D_2}$
Substituting the given value in above equation,
$\Rightarrow$ \[\dfrac{{g \times {3^2}}}{2} = \dfrac{{g \times {{\left( {t - 1} \right)}^2}}}{2} - \dfrac{{g \times {t^2}}}{2}\]
Cancel common factor in above equation,
We get,
$\Rightarrow$ ${3^2} = {t^2} - {\left( {t - 1} \right)^2}$
On simplifying,
$\Rightarrow$ $9 = {t^2} - \left( {{t^2} - 2t + 1} \right)$
Then, Here,
$\Rightarrow$ $9 = {t^2} - {t^2} + 2t - 1$
Solving the above equation,
We get,
$\Rightarrow$ $9 = 2t - 1$
Rearranging the given equation,
$\Rightarrow$ $2t = 10$
On simplifying,
We get,
$\Rightarrow$ \[t = 5\,s\]
Hence, The body has fallen for a time of $5$ second.

Thus, option B is the correct answer.

Note: A significant notion is the free-fall body principle. By applying it to distinct laws such as Newton's law of motion, students need to learn more about free fall bodies. It can be used to measure the velocity of the object after falling from rest after some specified amount of time. From weightlessness to free fall, visitors should not be confused.