How long does the body fall if in the last two seconds it has passed 60m?

Let’s start with the fact that we remember that the fall of bodies to the ground is a uniformly accelerated motion with an acceleration of gravity g = 10 m / s ^ 2.

Let us recall the formula for calculating the distance traveled with uniformly accelerated motion:

S = V1 * t + 0.5g * t ^ 2.

Let us express the initial velocity from the formula:

V1 = (S – 0.5 * g * t ^ 2): t = (60 – 0.5 * 10 * 2 ^ 2): 2 = 40: 2 = 20 m / s.

This means that from a height of 60 meters, the body began to fall at a speed of 20 m / s.

Let us recall the formula for calculating the speed at uniformly accelerated motion:

V2 = V1 + g * t.

So at the end of the movement, the speed is equal to:

V2 = 20 + 10 * 2 = 40 m / s.

Let us take V0 = 1 at the beginning of the fall and get:

V1 = V0 + 10t.

20 = 0 + 10t.

Hence, t = 20: 10 = 2 s.

The time from the beginning of the fall to the end has passed:

2 s + 2 s = 4 s.