2. An individual started standing on a force plate. He performed a countermoveme

Question

2. An individual started standing on a force plate. He performed a countermovement jump and then landed back to the force plate. Below is the hypothetical vertical ground reaction force graph. a. Identify when he started the countermovement jump as event A, when he left the force plate as event B, when he landed back to the force plate as event C b Calculate the jump height based on the vertical ground reaction force graph. (g -9.8 m/s3). 6 points 4500 4000 53500 3000 2500 2000 1500 1000 500 0 98 0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1 1.1 1.2 1.3 1.4 1.5 Time (s)

Explanation / Answer

2. During counterjump, reaction force (Normal force) increases while moving downwards and it is zero in the flight.

a) Time for Event A Ta = 0.4 s

    Time for Event B Tb = 0.9 s

    Time for Event C Tc = 1.4 s

b) Time of Flight = Tc - Tb

                            = 0.5 s

      Time of Ascent = (1/2)Time of Flight (Assuming no drag force acts on the body)

                               = 0.25 s

       Using Kinematic equation,

        s = vt - (1/2)at2      (s = vertical displacement, v=final velocity)

        For Maximum Height, s=H , v = 0 , t= time of ascent

        H = (1/2)(9.8)(0.252)

           = 0.3 m

           

       

Related Questions

Navigate

• Browse (All)
• Browse 2
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.