The stone falls freely, breaking away from the rock. The path traversed by us in the second second is?

To answer the question of our task, we need to find the difference in the distance traveled by the stone between the first and second seconds of its fall.

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

Substitute the known values ​​into the above formula (without initial speed) –

S = (gt²) / 2, we get –

S in the first second = (10 m / s² * (1 s) ²) / 2 = 5 m.

S in the second sec = (10 m / s² * (2 s) ²) / 2 = 20 m.

Therefore, in the second second, the body will pass 15 meters (20 m – 5 m) under the conditions of our example.