Ratio of distances covered by object falling freely under gravity in `1 ^(st),2^(nd)&3^(rd)` second 18 ........

Options

A. `1:3:5`
B. `1:2:5`
C. `1:4:9`
D. `1:5:9`

Answer

A

Explanation

Distance covered in `n^(th)` second,
In `d _(n) = v _(0)t + a/2 (2n -1) , v_(0) =0,`
`d_(n) = (a)/(2) (2n -1)`
`therefore ` For `n =1, d _(1) = g /2 (2xx 1-1) = g/2`
For `n = 2, d_(2) = g/2 (2 xx 2-1) = (3g)/(2)`
For `n =3, d _(3) = g/2) (2 xx 3-1) = (5g)/(2)`
`therefore d _(1) : d_(2) :d _(3) = (g)/(2) : (3g)/(2) : (5g)/(2) = 1:3:5`