What is the path of a freely falling body in 3 seconds?

The path traversed by the body during uniformly accelerated motion is determined by the formula S = v start * t + (at²) / 2, where S is the path, distance, v start is the initial speed, a is acceleration, t is time.

In our case, we have a free fall, that is, a – acceleration will be equal to acceleration free fall – g (let’s round its value up to 10 m / s²). And the initial speed, judging by our condition, is absent.

Substitute the known values ​​into the above formula – S = (gt²) / 2, we get – (10 m / s² * (3 s) ²) / 2 = 90 m / 2 = 45 m.

Answer: the path covered by the falling body in 3 seconds will be 45 meters.