Question

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.

Answer 1

Gravity is 9.8 m/s² means means every second distance travelled
increases by the distance in the previous second plus an extra 9.8m
during last second it fell 38m
previous second dist = 38 - 9.8m = 28.2
previous second = 28.2 - 9.8m = 18.4m
distance left = 18.4 - 9.8m = 8.6m
(so actually less than a second as it only travelled 8.6m)
total distance h = 38 + 28.2 + 18.4 + 8.6 = 93.2m

hope this is what is required

Answer 2

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  .