A 0 . 4kg mass takes 8s to reach ground when dropped from a certain...

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

Here, displacement is 8^th second is

${S}_{8} = 0 + \dfrac{1}{2} \times 10 \times ( 2 \times 8 - 1 )$

$⇒ {S}_{8} = 75m$

Loss of potential energy is

$\DeltaU = mg{S}_{8} = 0 . 4 \times 10 \times 75$

$⇒ \DeltaU = 300J$.