A body is dropped from certain height above the ground and covers `((5)/(36))^("th")` of the total height during the last second of the fall. Calculate height from which the body was dropped. `(g=10ms^(-2))`

Explanation

`"Given "S_(""nth")=(5)/(36)h,`
For a free fall, `h=(1)/(2)gn^(2)`
We know that `s^("nth")=v_(0)+(g)/(2)(2n-1)`
for a free fall `v_(0)=0`
`"i.e. "(5)/(36)h=(g)/(2)(2n-1)`
Substituting for h we get `(5)/(36)((1)/(2)gn^(2))=(g)/(2)(2n-1)`
`"i.e. "(5)/(36)n^(2)=2n-1" "5n^(2)=72n-36`
`"or "5n^(2)-72n+36=0`
`therefore" "n=(72pm sqrt((72)^(2)-4(5)(36)))/(2xx5)=(72pm66.8)/(10)=13.9s`
using the value of n in h `=1//2 gn^(2)`
We get `h=1//2xx10xx13.9xx13.9=966.0m`