A free-falling body passed 4/5 of its path at the last second of its fall. Find the time and height of the body fall?

Given:

g = 10 m / s2 (meters per second squared) – free fall acceleration;

S1 = S * 4/5 – the distance traveled by the body in the last second of the fall.

It is required to determine t (seconds) – time and S (meter) – fall height.

By the condition of the problem, the body falls freely, that is, without initial velocity. Then:

S = g * t ^ 2/2.

During the last second of the fall (t1 = 1 second), the body has passed a distance equal to:

S1 = g * t ^ 2/2 – g * (t-1) ^ 2/2, or

4 * S / 5 = g * t ^ 2/2 – g * (t-1) ^ 2/2;

2 * t ^ 2/5 = t ^ 2/2 – (t-1) ^ 2/2;

2 * t ^ 2/5 = (t ^ 2 – (t ^ 2 – 2 * t + 1)) / 2;

2 * t ^ 2/5 = (t ^ 2 – t ^ 2 + 2 * t – 1) / 2;

2 * t ^ 2/5 = (2 * t – 1) / 2;

4 * t ^ 2 = 10 * t – 5;

4 * t ^ 2 – 10 * t + 5 = 0.

We got the following quadratic equation:

4 * t ^ 2 – 10 * t + 5 = 0;

D = 100 – 4 * 4 * 5 = 100 – 80 = 20; D ^ 0.5 = 20 ^ 0.5 = 4.5.

t1 = (10 + 4.5) / 8 = 14.5 / 8 = 1.8 seconds;

t2 = (10 – 4.5) / 8 = 5.5 / 8 = 0.7 seconds (does not match the problem statement).

Then:

S = g * t1 ^ 2/2 = 10 * 1.8 ^ 2/2 = 5 * 1.8 ^ 2 = 16.2 meters.

Answer: the body fell from a height of 16.2 meters for 1.8 seconds.