What is the value of the rate of fall of a 200 kg load that has fallen from a height of 2 m?

Task data: m (weight of the dropped load) = 200 kg; h (drop height) = 2 m.

Reference values: g (acceleration due to gravity, load) ≈ 10 m / s2.

We express the value of the speed of the load under consideration at the end of the fall from the formula: S = h = (V ^ 2 – V0 ^ 2) / 2а = (V ^ 2 – 0 ^ 2) / 2g = V ^ 2 / 2g, whence V = √ ( 2g * h).

Let’s calculate: V = √ (2 * 10 * 2) = 6.32 m / s.

Answer: The load under consideration had a speed of 6.32 m / s at the end of the fall.