The body falls freely from a height of 180 m. Divide this distance into such sections

The body falls freely from a height of 180 m. Divide this distance into such sections so that the body passes them in equal periods of time.

Data: h – the height of the fall of the specified body (h = 180 m).

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

1) Duration of the fall: t = √ (2 * S / a) = √ (2 * h / g) = √ (2 * 180/10) = 6 s.

2a) Time intervals t = 1 s: h1 = g * t1 ^ 2/2 = 10 * 1 ^ 2/2 = 5 m;

h2 = h (2) – h1 = g * t2 ^ 2/2 – h1 = 10 * 2 ^ 2/2 – 5 = 15 m;

h3 = h (3) – h1,2 = g * t3 ^ 2/2 – h1,2 = 10 * 3 ^ 2/2 – 20 = 25 m;

h4 = h (4) – h1,2,3 = g * t4 ^ 2/2 – h1,2,3 = 10 * 4 ^ 2/2 – 45 = 35 m;

h5 = h (5) – h1,2,3,4 = g * t5 ^ 2/2 – h1,2,3,4 = 10 * 52/2 – 80 = 45 m;

h6 = h (6) – h1,2,3,4,5 = g * t62 / 2 – h1,2,3,4,5 = 10 * 62/2 – 125 = 55 m.

2b) Segments of time t = 2 s: h1 = 5 + 15 = 20 m;

h2 = 25 + 35 = 60 m;

h3 = 45 + 55 = 100 m.

2c) Segments of time t = 3 s: h1 = 5 + 15 + 25 = 45 m;

h2 = 35 + 45 + 55 = 135 m.