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Question: What will be the ratio of the distance covered by a freely falling body from rest in \({4^{th}}\) an...

What will be the ratio of the distance covered by a freely falling body from rest in 4th{4^{th}} and 5th{5^{th}} second of journey?

Explanation

Solution

Let us first know about free fall. A free falling object is one that is falling only due to gravity's impact. A state of free fall is defined as an object that is only affected by the force of gravity. There are two important motion properties that free-falling objects possess: air resistance does not exist for free-falling objects. On Earth, all free-falling objects accelerate downward at a rate of 9.89.8metres per second.

Formula Used:
Distance covered in nth{n^{th}} second is given by
Sn=u+a2(2n−1){S_n} = u + \dfrac{a}{2}(2n - 1)
Where, u=u = Initial velocity, a=a = Acceleration due to gravity and n=n = Number of seconds.

Complete step by step answer:
In the technical definition of the term "free fall," an item may or may not be falling down in the common sense. Although an object going upwards is not typically thought to be falling, it is said to be in free fall if it is merely subjected to the force of gravity. The Moon is thus in free fall around the Earth, despite the fact that its orbital speed keeps it at a great distance from the planet's surface.

So as we discussed above:
Distance covered in nth{n^{th}} second is given by
Sn=u+a2(2n−1){S_n} = u + \dfrac{a}{2}(2n - 1)
⇒u=0,a=g\Rightarrow u = 0,a = g
When the number of seconds is n=4n = 4
S4=g2(2×4−1)=7g2{S_4} = \dfrac{g}{2}(2 \times 4 - 1) = \dfrac{{7g}}{2}
When the number of seconds is n=5n = 5
S5=g2(2×5−1)=9g2{S_5} = \dfrac{g}{2}(2 \times 5 - 1) = \dfrac{{9g}}{2}
Now let us find the ratio:
∴S4S5=79\therefore \dfrac{{{S_4}}}{{{S_5}}} = \dfrac{7}{9}

So, the ratio of the distance covered by the freely falling body is 7:97:9.

Note: Let us know a little more about free fall. The term "free fall" is frequently used in a broader sense than it is strictly defined. Free fall is the term used to describe falling through the atmosphere without the use of a parachute or other lifting device. Because the aerodynamic drag forces prohibit full weightlessness in such scenarios, a skydiver's "free fall" after achieving terminal velocity gives the impression that the body's weight is supported on a cushion of air.