The body falls freely from a height of 80 m. What path does it travel in the last second of the fall?

Body path when falling:
S = V0 * t + a * t² / 2, where S is the height of the body’s fall (S = 80 m), V0 is the initial velocity of the body (since the body falls freely, V0 = 0 m / s), t is the time of the fall , a – acceleration of the body (a = g = 10 m / s²).
S = V0 * t + a * t² / 2 = g * t² / 2.
t = sqrt (2S / g) = sqrt (2 * 80/10) = 4 s.
S = S1 + S2, S1 is the path covered in 3 seconds of the fall, S1 is the path in the last second of the fall.
S1 = g * t² / 2 = 10 * 3² / 2 = 45 m.
S2 = S – S1 = 80 – 45 = 35 m.
Answer: The body has passed 35 m.