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.

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

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.