The body falls freely without initial speed from a height of 45 m.

The body falls freely without initial speed from a height of 45 m. How many times the distance traveled in the last second of movement is more than in the first.

Task data: h (fall height) = 45 m.

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

1) Total duration of the fall: t = √ (2h / g) = √ (2 * 45/10) = 3 s.

2) Path in the first second: h1 = g * 1 ^ 2/2 = 10 * 1 ^ 2/2 = 5 m.

3) Path 1 s before the fall: h2 = g * (t – 1) ^ 2/2 = 10 * (3 – 1) ^ 2/2 = 20 m.

4) Path in the last second: h3 = h – h2 = 45 – 20 = 25 m.

5) Ratio of distances covered: k = h3 / h1 = 25/5 = 5 p.

Answer: The path in the last second is 5 times longer.