1.

A body is dropped from a height of 125 m. If = 10 ms^(-2) what is the ratio of the distances travelled by it during the first and last second of its motion :

Answer»

`2:9`
`1:9`
`1:3`
`4:9`

Solution :`T=SQRT(2H)/(g)=sqrt((2xx125)/(10))=5s`
During first SECOND the body falls through a height
`h_(1)=(1)/(2)g t_(1)^(2)=(1)/(2)xx101^(2)=5M`
During first four second the body falls through height
`h_2=(1)/(2) "gt"_(2)^(2)=(1)/(2)xx10xx4^(2)=80 m`
`:.` height through which the body falls in 5TH second is `h_(3)`=125-80=45 m
`:.` `(h_(1))/(h_3)=(5)/(45)=(1)/(9)`


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