(c) As shown in adjoining figure ball is falling from height 2$h$ and rebounding to a height $h$ only. It means that velocity of ball jus before collision
$u = \sqrt{\frac{2(2h)}{g}} = \sqrt{\frac{4h}{g}}$ and velocity just after collision
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$v = - \sqrt{\frac{2h}{g}}$
$\therefore e = \frac{- v\sqrt{\frac{2h}{g}}}{u\ \sqrt{\frac{4h}{g}\ }} = \frac{1}{\sqrt{2}}$