The height of the fall of the object h = 45m, on the second half of the road

The height of the fall of the object h = 45m, on the second half of the road Vaverage =? V0 = 0 disregarding wind resistance (Vaverage ~ 25.9 m / s).

h = 45 m.

V0 = 0.

g = 10 m / s2.

Vср2 -?

To find the average speed of movement Vav2 on the second half of the path, it is necessary to divide the half of the traversed path h2 by the time of its passage t2: Vav2 = h2 / t2.

h2 = h / 2.

h1 = h / 2.

Let us find the travel time of the second half of the path t2 by the formula: t2 = t – t1, where t is the travel time of the entire path, t1 is the travel time of the first half of the path.

h = g * t ^ 2/2.

t = √ (2 * h / g).

t = √ (2 * 45 m / 10 m / s2) = 3 s.

h1 = g * t1 ^ 2/2.

t1 = √ (2 * h1 / g) = √ (2 * h / 2 * g) = √ (h / g).

t1 = √ (45 m / 10 m / s2) = 2.1 s.

t2 = 3 s – 2.1 s = 0.9 s.

Vav2 = 22.5 m / 0.9 s = 25 m / s.

Answer: the average speed of movement of the object in the second half of the path was Vav2 = 25 m / s.