With a free fall of a stone from a certain height, its average speed along the entire path was = 40 m / s.

With a free fall of a stone from a certain height, its average speed along the entire path was = 40 m / s. Determine the average speed of the stone for the second half of the fall time.

Given:

v = 40 m / s – average speed during free fall of the stone;

g = 10 m / s ^ 2 – acceleration of gravity.

It is required to find v1 – the average speed of the stone for the second half of the fall time.

According to the condition of the problem, the stone falls without initial speed.

The average speed over the entire route is:

v = H / t = g * t ^ 2/2 * t = g * t / 2, hence the total fall time is:

t = 2 * v / g = 2 * 40/10 = 80/10 = 8 seconds.

Then the distance from which the stone fell is equal to:

H = g * t ^ 2/2 = 10 * 8 * 8/2 = 5 * 8 * 8 = 320 meters.

In the first half of the time, t1 = t / 2 = 4 seconds, the stone will cover the distance:

H1 = g * t1 ^ 2/2 = 10 * 4 * 4/2 = 5 * 4 * 4 = 80 meters.

Accordingly, in the second half of the time, the stone will cover the distance:

H2 = H – H1 = 320 – 80 = 240 meters.

Then, the average speed for the second half of the time is:

v1 = H2 / t1 = 240/4 = 60 m / s.

Answer: the average speed for the second half of the time is 60 m / s.