A body falling freely descends 0.3m in 0.1 s and 0.398 m in the next 0.1 s in some other planet . Calculate the value of g in that planet ?

Explanation

Given : During `t_(1) = 0.1` distance covered is `s_(1) = 0 . 3` m
During next `t_(2) = 0` is, distance covered is `s_(2) = 0 . 398 m`
We know `s = ut +(1)/(2) at^(2)`
In this case `u = 0, a = g, s = (1)/(2) g t^(2)`
`t = t_(1) + t_(2) = 0 . 1 + 0 . 1 = 0 . 2`
`:. g = (2s)/(t^(2))`
For this problem
`g = (2(s_(2)- s_(1)))/(t^(2))`
`g = (2 (0. 398 - 0 . 3))/((0 . 2)^(2))`
`= (2 xx (0. 098))/(0 . 0 4) = (0. 196)/(0 . 0 4)`
` g = (19.6)/(4) =4 . 9 m//s^(2)`
`:. ` Acceleration due to gravity in that planet `= 4 . 9 m//s^(2)`