F (kN)1.2-1.0-0.8-0.6 F0.4-0.2-TAKEOFFLANDINGA₁A3At20.0-0.2-Atı!Atз-0.4-A2-0.6--0.8-F2AIRBORNE-1.0-Consider the force versus time recording shown in Fig. 11.9 for an athlete makingvertical jumps on a force platform. The force scale isnormalized with the weight of the athlete so that theforce reading is zero when the athlete is stationary(standing still or crouching). The reason for normalizingthe force measurement is to be able to disregard theeffect of gravitational acceleration on the impulsecalculations. In this case, a positive force means a forceexerted on the platform due to factors other than theweight of the person. The force versus time graph hasthree distinct regions. An initial “takeoff push" duringwhich the athlete exerts a positive force on the platform,an "airborne" region during which the athlete is not incontact with the platform, and a "landing” period inwhich the athlete again exerts impulsive forces on theplatform. These regions are approximated withrectangular areas A1 and A2, and a triangular area A3.Boundaries of these regions are shown with dashed linesin Fig. 11.9. The approximate force applied and the duration of takeoff push are about F1= 600 N and At₁ = 0.3 s. The force reading is about F2 = -800 N while the athlete isairborne, and the athlete remains in the air for about At2 = 0.4 s. This suggests that theweight of the athlete is about 800 N. Therefore, the athlete has a mass of about m =(800 N)/(9.8 m/s²) = 82 kg. The maximum impulsive force the athlete exerts on theplatform during landing is about F3 = 1000 N which reduces to zero (in the normalizedscale) in a time interval of about At3 = 0.4 s.-1.2Fig. 11.9 Normalized force versustime plot for the athlete(1 kN = 1000 N)Determine an approximate takeoff velocity of the center of gravity of the athlete,calculate the height of jump, and determine the impulse and momentum of the athleteduring landing by using the approximated areas under the force versus time curve.

Question

F (kN)1.2-1.0-0.8-0.6 F0.4-0.2-TAKEOFFLANDINGA₁A3At20.0-0.2-Atı!Atз-0.4-A2-0.6--0.8-F2AIRBORNE-1.0-Consider the force versus time recording shown in Fig. 11.9 for an athlete makingvertical jumps on a force platform. The force scale isnormalized with the weight of the athlete so that theforce reading is zero when the athlete is stationary(standing still or crouching). The reason for normalizingthe force measurement is to be able to disregard theeffect of gravitational acceleration on the impulsecalculations. In this case, a positive force means a forceexerted on the platform due to factors other than theweight of the person. The force versus time graph hasthree distinct regions. An initial “takeoff push&#34; duringwhich the athlete exerts a positive force on the platform,an &#34;airborne&#34; region during which the athlete is not incontact with the platform, and a &#34;landing” period inwhich the athlete again exerts impulsive forces on theplatform. These regions are approximated withrectangular areas A1 and A2, and a triangular area A3.Boundaries of these regions are shown with dashed linesin Fig. 11.9. The approximate force applied and the duration of takeoff push are about F1= 600 N and At₁ = 0.3 s. The force reading is about F2 = -800 N while the athlete isairborne, and the athlete remains in the air for about At2 = 0.4 s. This suggests that theweight of the athlete is about 800 N. Therefore, the athlete has a mass of about m =(800 N)/(9.8 m/s²) = 82 kg. The maximum impulsive force the athlete exerts on theplatform during landing is about F3 = 1000 N which reduces to zero (in the normalizedscale) in a time interval of about At3 = 0.4 s.-1.2Fig. 11.9 Normalized force versustime plot for the athlete(1 kN = 1000 N)Determine an approximate takeoff velocity of the center of gravity of the athlete,calculate the height of jump, and determine the impulse and momentum of the athleteduring landing by using the approximated areas under the force versus time curve.

Accepted Answer

GivenConsider the force versus time recording shown in Fig. 11.9 for an athlete making vertical…