Question

A body is falling freely under gravity. The distance covered by in first, second and third second of its motion are in ratio

Answer 1

1:3:5:7:9......
Use S = (g/2) (2t-1)

Answer 2

The ratio of "first , second, and third motion is in the ratio of $\bold{frac{g}{2} : 2 g : \frac{9 g}{2}=1 : 4 : 9}$

Solution:

According to the given question here we need to find the ration of body at time t = 1s, t = 2s and t = 3 s.

According to Newton’s equation of motion,

we are having  that,

$s=u t+\frac{1}{2} a t^{2}$

Where “s” is the distance travelled by body , “u” is the initial speed and  “a” is the acceleration .

Here the body is falling freely due to gravity, so the acceleration will be that due to gravity.

a = g and the initial speed of the body will be 0.

For distance travelled in t = 1s  

$s=0 \times 1+\frac{1}{2} g 1^{2}=\frac{g}{2}$

For distance travelled in t = 2s

$s=0 \times 2+\frac{1}{2} g 2^{2}=2 g$

For distance travelled in t = 3s.

$s=0 \times 3+\frac{1}{2} g 3^{2}=\frac{9 g}{2}$

The ratio of the distance travelled in t = 1s , t= 2s and t = 3s is

$\frac{g}{2} : 2 g : \frac{9 g}{2}=1 : 4 : 9$