An athlete weighing 80 kg jumps into the pool from a height of 5 m. In the water, he dives

An athlete weighing 80 kg jumps into the pool from a height of 5 m. In the water, he dives to the maximum depth in 0.4 s. With what average force does the water act on the athlete?

m = 80 kg.

h = 5 m.

g = 9.8 m / s ^ 2.

t = 0.4 s.

Fср -?

Let us find the speed of an athlete during contact with water according to the law of conservation of energy.

m * h * g = m * V ^ 2/2.

V = √ (2 * h * g).

Find the athlete’s acceleration in the water.

a = V / t = √ (2 * h * g) / t.

Under water, it moves under the action of gravity m * g and resistance force Fav.

Fav – m * g = m * a.

Fav = m * g + m * a = m * (g + √ (2 * h * g) / t).

Fav = 80 kg * (9.8 m / s ^ 2 + √ (2 * 5 m * 9.8 m / s ^ 2) / 0.4 s) = 2764 N.

Answer: the average force of water resistance is Fav = 2764 N.