H29 X V fx C D E G H Time (s) Vertical Ground Reaction Force (N) Area 0 822.76 0.OOO5 823.12 0.4114709 0.001 822.94 0.4115165 0.0015 822.94 0.4114709 0.002 821.85 0.411197 0.0025 822.94 0.411197 Vertical Ground Reaction Force 0.003 822.7 0.4114252 35 00 0.0035 823.12 0.4114709 Z 30 00 0.004 823.31 0.4116079 0.0045 823.12 0.4116079 0.005 822.94 0.4115165 20 00 0.005S 822.76 0.4114252 15 00 0.006 823.12 0.4114709 823.31 0.4116079 10 00 0.0065 Vertica; Ground 0.007 822.21 0.4113796 500 0.0075 823.49 0.4114253 0 821.66 0.4112883 2000 40:00 60 00 80 00 10000 12000 0.008 0.0085 822.58 0.41106 Time (s) 0.009 822.58 0.4112883 0.0095 822.03 0.4111513 0.01 820.75 0.4106947 0.0105 823.12 0.4109687 0.011 823.49 0.4116535 0.0115 822.58 0.4115166 0.012 822.76 0.4113339 0.0125 822.94 0.4114252 0.013 824.40 0.4118361 0.0135 822.76 0.4117905 0.014 821.48 0.41106 0.0145 821.66 0.410786 0.015 821.48 0.410786 0.0155 822.39 0.4109687 0.016 821.66 0.4110143 0.0165 820.75 0.4106034 0.017 823.12 0.4109687 0.0175 822.76 0.4114709 0.018 823.12 0.4114709 0.018 822.21 0.4113339 0.019 822.94 0.4112883 0.0195 823.31 0.4115622 0.02 822.76 0.4115166 0.0205 822.94 0.4114252 0.021 822.58 0.4113796 0.0215 822.21 0.411197 0.022 823.85 0.4115166 0.0225 822.39 0.4115622 0.023 823.31 0.4114253 0.0235 822.58 0.4114709 0.024 822.76 0.4113339 0.0245 822.76 0.4113796 0.025 822.58 0.4113339 0.0255 822.76 0.4113339 0.026 822.94 0.4114252 0.0265 823.12 0.4115165 0.027 322.76 0.4114709 0.0275 822.76 0.4113796 0.028 323.12 0.4114709 0.0285 823,49 0.4116535 0.029 822.76 0.4115622 hard_1 Ready Accessibility: InvestigateDraw with Trackpad STEPS 1. Download the "Jumpforce.xls" data sheet. This file contains the vertical ground reaction force (as measured by a force plate) during a counter movement jump. A counter movement jump and the associated ground reaction force is shown below (from Linthorne et al., 2001) (a) a d 8 h Vertical Ground Reaction Force (N) Time (s) Focus MacBook AirANALYSES AND QUESTIONS 1. How long is the jumper in the air? (vertical jump only; no friction) (J= impulse) 2. Draw the free body diagram of the jumper when he is standing on the ground. Remember to label your forces clearly 3. What is the weight of the individual? (assume that the person is static and not moving at the start of the data collection). Explain your reasoning in terms of the free body diagram and Newton's laws 4. Compute the maximum 'upward' acceleration of the jumper prior to the point where he leaves the ground. 1606 5. Draw the free body diagram of the jumper when he is in the air. 6. Since you know the time that the jumper is in the air, use projectile motion to compute the maximum height reached by the jumper's CoG. Assume the initial height of the Center of gravity at take-off = 1.2 m, and that the landing height is equal to the take-off height 7. What is the net impulse of the jump prior to the point when the jumper leaves the ground? First, use the trapezoidal rule in Excel to compute the area under the force-time curve (remember to only calculate the area until the time the person leaves the ground). This area will be the impulse due to the ground reaction force. 1130.283 From this ground reaction force impulse, subtract the impulse due to body weight as well to find the net impulse prior to takeoff. The impulse due to body weight is a lot simpler to calculate since body weight does not change (so you can simply use Impulse = F x time that force is applied) 8. Use impulse-momentum equation to compute the maximum height reached by the jumpers' CoG. Assume the initial height of the Center of gravity at take-off = 1.2 m. Do you think this method will be more/less accurate than the projectile method (think about which method you had to make more assumptions - that will generally be the less accurate method)? Express the peak landing force as a % of body weight (%BW = Peak landing Force/Body weight * 100). Why might it be useful to express this landing force in terms of % body weight (instead of directly in Newtons)? Peak landing force 3000 10. Describe one thing you could do to reduce the landing force during the jump? Explain your reasoning in terms of mechanical principles. d States) Focus MacBook Air 80 DOO DDD DD F3 F4 F5 F6 F7 F8 F9 F10 % &