Question

The standing vertical jump is a good test of an athlete's strength and fitness. The athlete goes into a deep crouch, then extends theirlegs rapidly; when theirlegs are fully extended, theyleave the ground and rise to theirhighest height.It is the force of the ground on the athlete during the extension phase that accelerates the athlete to the final speed with which theyleave the ground.A good jumper can exert a force on the ground equal to twice theirweight. If thecrouch is 65 cm deep, how far off the ground do theyrise

Answer 1

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 = $\sqrt{4gx}$

     

let's calculate

          v = $\sqrt{ 4 \ 9.8 \ 0.65}$

          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  $\frac{v^2}{g}$

let's calculate

           h = ½ 5.05²/9.8

           h = 1.3 m