The body falls from a height of 125 m. To determine the path of the body in the last 3 seconds.

Initial data: h (total fall height) = 125 m; t2 (considered time) = 3 s.

Constants: g (acceleration due to gravity) ≈ 10 m / s2.

1) Total fall time: h = g * t ^ 2/2, whence t = √ (2h / g) = √ (2 * 125/10) = 5 s.

2) The first time of the fall: t1 = t – t ^ 2 = 5 – 3 = 2 s.

3) Distance for the first 2 s of flight: h1 = g * 1/2 = 10 * 2 ^ 2/2 = 20 m.

4) Track in the last 3 s: h2 = h – h1 = 125 – 20 = 105 m.

Check: h2 = V1 * t2 + g * t2 ^ 2/2 = g1 * t1 * t2 + g * t2 ^ 2/2 = 10 * 2 * 3 + 10 * 3 ^ 2/2 = 60 + 45 = 105 m (right).

Answer: During the last 3 s of the fall, the body will fly 105 m.