An athlete weighing 80 kg jumps into the pool from a height of 5 m. In water, it dives to its maximum depth in 0.4 s.

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

These tasks: m (the mass of the athlete who jumped into the pool) = 80 kg; h (height at which the athlete was) = ​​5 m; t (duration of immersion at maximum depth) = 0.4 s.

Constants: g (acceleration due to gravity) ≈ 10 m / s2.

1) Speed ​​before immersion in water: Ek = Ep and m * V2 / 2 = m * g * h, whence we express: V = √ (g * h * 2) = √ (10 * 5 * 2) = 10 m / from.

2) Acceleration of an athlete when immersed in water: V = V0 + a * t = 0, whence we express: a = V / t = 10 / 0.4 = 25 m / s2.

3) Average force of water resistance to the athlete’s immersion: m * a = Fcopr. – Fт, whence we express: Fcopr. = m * a + Ft = m * a + m * g = m * (a + g) = 80 * (25 + 10) = 2.8 * 10 ^ 3 H = 2.8 kN.