A 0 . 4   kg mass takes 8   s to reach ground when dropped from a certain height P above surface…

Displacement in ${\mathrm{n}}^{\mathrm{th}}$ second is given by $S_n=u+\frac{a}{2}\left(2n-1\right)$

Here, displacement is 8^th second is

$S_8=0+\frac{1}{2}\times 10\times (2\times 8-1)$

$\Rightarrow S_8=75 \mathrm{m}$

Loss of potential energy is 

$$\begin{gathered} \Delta U=mg{S}_8 \\ =0.4\times 10\times 75 \end{gathered}$$

$\Rightarrow \Delta U=300 \mathrm{J}$.