At the last second of free fall, the body passed a path n = 2 times longer than the previous one.

At the last second of free fall, the body passed a path n = 2 times longer than the previous one. Find the total fall time t if the initial velocity of the body is zero.

Task data: S1 (path in the last second) = 2S2 (path in the penultimate second).

1) Path at the last second: S1 = g * t ^ 2/2 – g * (t – 1) ^ 2/2.

2) Path in the penultimate second: S2 = g * (t – 1) ^ 2/2 – g * (t – 2) ^ 2/2.

3) The resulting equality: g * t ^ 2/2 – g * (t – 1) ^ 2/2 = 2 * (g * (t – 1) ^ 2/2 – g * (t – 2) ^ 2 / 2).

t ^ 2/2 – (t – 1) ^ 2/2 = (t – 1) ^ 2 – (t – 2) ^ 2.

t ^ 2/2 – (t ^ 2 – 2t + 1 ^ 2) / 2 = (t ^ 2 – 2t + 1 ^ 2) – (t ^ 2 – 4t + 2 ^ 2).

0.5t ^ 2 – 0.5t ^ 2 + t – 0.5 = t ^ 2 – 2t + 1 – t ^ 2 + 4t – 4.

t – 0.5 = 2t – 3.

t = 2.5 s.

Answer: The total fall time is 2.5 s.