Question

find the ratio of the distances travelled by the freely falling body in 1st, 3rd and 5th seconds​

Answer 1

Answer:

Explanation:

The ratio is 1:3:5.

[Read this if you want to see the formulas.

Using the two equations for constant acceleration:

1. s=ut+12at2

2. v=u+at

In the first second, u=0, t=1 s, a=−g, so s=−g/2 and v=−g.

In the second second, u=−g, t=1 s, a=−g, so s=−g−g/2=−3g/2 and v=−g−g=−2g.

In the third second, u=−2g, t=1 s, a=−g, so s=−2g−g/2=−5g/2 and v=−2g−g=−3g.

So the ratio of displacements is −g/2:−3g/2:−5g/2 or simply 1:3:5.]

Answer 2

1:3:5 will be the ratio for the following