The distance travelled by a body falling freely from rest in first, second and thirdseconds are in the ratio

Let's assume that the distance travelled by the body falling freely from rest in the first second is represented by "d1", in the second second is represented by "d2", and in the third second is represented by "d3".

According to the laws of motion, the distance travelled by a freely falling body can be calculated using the equation:

d = (1/2) * g * t^2

where d is the distance, g is the acceleration due to gravity, and t is the time.

In the first second, the time (t) is 1 second. Therefore, the distance travelled in the first second (d1) can be calculated as:

d1 = (1/2) * g * (1^2) = (1/2) * g

In the second second, the time (t) is 2 seconds. Therefore, the distance travelled in the second second (d2) can be calculated as:

d2 = (1/2) * g * (2^2) = 2 * g

In the third second, the time (t) is 3 seconds. Therefore, the distance travelled in the third second (d3) can be calculated as:

d3 = (1/2) * g * (3^2) = (9/2) * g

Now, let's find the ratio of the distances travelled in the first, second, and third seconds:

d1 : d2 : d3 = (1/2) * g : 2 * g : (9/2) * g

To simplify the ratio, we can cancel out the common factor of g:

d1 : d2 : d3 = 1/2 : 2 : 9/2

To further simplify the ratio, we can multiply each term by 2:

d1 : d2 : d3 = 1 : 4 : 9/2

Therefore, the ratio of the distances travelled by the body falling freely from rest in the first, second, and third seconds is 1 : 4 : 9/2.