Prove that,a freely falling body of mass m loses its potential energy by `1/2 mg^2(2t-1)` during its fall in the t th second.

Explanation

Suppose the body starts from rest. The displacement in t seconds,
`x=1/2 "gt"^2`.
The displacement in (t-1) seconds,
`y=1/2 g(t-1)^2`
`therefore` Displacement in the t th second ,
`h=x-y=1/2 "gt"^2-1/2g(t-1)=1/2g(2t-1)`
`therefore` Decrease in potential energy in the t th second
`=mgh =mg*1/2g(2t-1)=1/2 mg^2(2t-1)`.