The safety of a falling person depends on the momentum with which he falls on a planet. Hence, lesser velocity is the limiting vector. Let $v$ be the velocity of fall. Then, \[v^2=2gh=2g’h’\] where, $g$ and $g’$ are the accelerations due to gravity on the earth and the planet and $h$ and $h’$ are the safe heights on the earth and the planet respectively. Thus, \[h’=\frac{gh}{g’}=\frac{9.8×2}{1.96}=10\;\text{m}\]

Hence, the safe height of fall on the planet is 10 m.