Answer:
 h = 1.3 m
Step-by-step 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 = 
 
 
 
let's calculate 
 v = 
 
 
 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 
 
 
 
let's calculate 
 h = ½ 5.05²/9.8 
 h = 1.3 m