Answer:
h = 1.3 m
Explanation:
For this exercise we can use the relationship between work and kinetic energy
W = Δx = K_f - K₀
in the exercise they indicate that the strength of the athlete is over twice his weight, therefore the ratio of the floor directed upwards has the same value
F = 2W
F = 2 mg
the displacement is x = 0.65 m, note that the direction of the force and the displacement is the same and the initial velocity is zero due to being crouched at rest, we substitute
F x = ½ m v² - 0
v² = 2 (2mg) x / m
v = [tex]\sqrt{4gx}[/tex]
let's calculate
v = [tex]\sqrt{ 4 \ 9.8 \ 0.65}[/tex]
v = 5.05 m / s
already in the air energy is conserved
starting point. Just when it comes off the ground
Em₀ = K = ½ m v²
final point. When is it at the highest part of the trajectory
Em_f = U = m g h
Em₀ = Em_f
½ m v² = m g h
h = ½ [tex]\frac{v^2}{g}[/tex]
let's calculate
h = ½ 5.05²/9.8
h = 1.3 m