Answer
The elevator strikes the ground with a speed of $~~48.5~m/s$
Work Step by Step
We can let "down" be the positive direction.
We can find the speed when the elevator strikes the ground:
$v^2 = v_0^2+2ay$
$v^2 = 0+2ay$
$v = \sqrt{2ay}$
$v = \sqrt{(2)(9.8~m/s^2)(120~m)}$
$v = 48.5~m/s$
The elevator strikes the ground with a speed of $~~48.5~m/s$
