Answer
$v_{B}=\sqrt {v^{2}_{0}+2gh}\left( 3\right) \approx 21\dfrac {m}{s}$
Work Step by Step
The total energy of snowball respect to the ground initially will be: $E_{tot}=E_{k0}+E_{p0}=\dfrac {mv^{2}_{0}}{2}+mgh(1)$ Total energy of snowball is conserved so when it hits the ground lets write total energy of snowball respect to ground again : $ E_{tot}=E_{kB}+E_{pB}=\dfrac {mv^{2}_{B}}{2}+0\left( 2\right) $ so from (1) and (2) we get: $v_{B}=\sqrt {v^{2}_{0}+2gh}\left( 3\right) \approx 21\dfrac {m}{s}$
