Question

The distance traveled by a body falling freely from rest in 2^nd, 3^rd and 5^th second of its motion is in the ratio of

• A. 7:5:3
• B. 3:5:9
• C. 5:3:7
• D. 5:7:3

Answer 1

The correct option is B

3:5:9

Step 1: Calculating distance covered by the body

1. We know the formula for calculating the distance covered by a body in a particular second that is given as$S_{nth}=u+\frac{a}{2}\left(2n-1\right)$.
2. In this equation $s$is the distance covered, $n$ is a particular time, $u$ is the initial velocity, and $a$is the acceleration of the body.
3. Here the body was initially at rest, $u$ will be zero
4. The body is falling freely under gravity. so acceleration will be due to gravity which means $a$ = $g$
5. So the equation will be $S_{nth}=0+\frac{g}{2}\left(2n-1\right)$
6. For the $2nd$second, $n=2$, the distance will be $S_{2nd}=\frac{g}{2}\left(2\times 2-1\right)$ $\Rightarrow S_{2nd}=\frac{3g}{2}$
7. For the $3rd$second, $n=3$, distance will be $S_{3rd}=\frac{g}{2}\left(2\times 3-1\right)$$\Rightarrow S_{3rd}=\frac{5g}{2}$
8. For the $5th$second, $n=5$and distance covered will be $S_{4th}=\frac{g}{2}\left(2\times 5-1\right)$$\Rightarrow S_{5th}=\frac{9g}{2}$

Step 2: Calculating the ratio of the three distances

1. The ratio of the distances covered by the freely falling body in$2nd$, $3rd$and $5th$seconds respectively is

$S_{2nd}:S_{3rd}:S_{5th}=\frac{3g}{2}:\frac{5g}{2}:\frac{9g}{2}$

$\Rightarrow S_{2nd}:S_{3rd}:S_{5th}=3:5:9$

Hence,

The ratio of the distances covered by the body is$3:5:9$.