Physics, asked by nidhanajmaranjith, 1 year ago

The distances traversed during equal interval of time by a freely falling body from rest are in the ratio?

Answers

Answered by alinakincsem
20

The ratio is 1:3:5.




To get the formulas, we say




Using the two equations for constant acceleration:


1. s=ut+12at2

2. v=u+at


According to the ratio:




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 displacement is −g/2:−3g/2:−5g/2 or simply put 1:3:5

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