An object is released from rest at a height h. during the final second of its fall, it traverses a distance of 38m. determine the value of h.

During the final second of its fall, it falls a distance of 38m.

Its average speed during that final second was 38 m/s.

But we know that it gained 9.8 m/s of speed during that second.

Its speed at the end of that second must have been (38+4.9) = 42.9 m/s ,
and at the beginning of that second must have been (38-4.9) = 33.1 m/s .

Since its speed at the beginning of the final second was  33.1 m/s,
that final second began at  (33.1/9.8) = 3.378 seconds after the drop.

All together, when the final second is added onto that, the object
fell for a total of  4.378 seconds.

  Distance of fall from rest = (1/2) (g) (t)²

                                      = (4.9 m/s²) (4.378 s)²

                                      = (4.9 m/s²) (19.163 s²)

                                      =      93.9  meters  .

Domain eukarya; kingdom plantae should be the answer.