I assume you mean 12 meters, not 12 kilometers
in general for constant acceleration a:
(the mass has nothing to do with this question)
v = Vi + a t
h = Hi + Vi t + (1/2) a t^2
here
v = - g t = -9.81 t
h = 12 - 4.9 t^2
so for v = 5 (actally-5, it goes down)
-5 = -9.81 t
t = 0.51 second
h = 12 - 4.9(0.51)^2=12 - 1.27 = 10.7 meters
a 2kg mass is dropped from 12km height at what height would the velocity be 5m/s?
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