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Your answer is partially correct. A student (m -63 kg) falls freely from rest and strikes the ground. During the collision with the ground, he comes to rest in a time of 0.04s. The average force exerted on him by the ground is +18000 N, where the upward direction is taken to be the positive direction. From what height did the student fall? Assume that the only force acting on him during the collision is that due to the ground. 6.7 m

Your answer is partially correct. A student (m -63 kg) falls freely from rest and strikes the ground. During the collision with the ground, he comes to rest in a time of 0.04s. The average force exerted on him by the ground is +18000 N, where the upward direction is taken to be the positive direction. From what height did the student fall? Assume that the only force acting on him during the collision is that due to the ground. 6.7 m

College Physics
College Physics
11th Edition
ISBN:9781305952300
Author:Raymond A. Serway, Chris Vuille
Publisher:Raymond A. Serway, Chris Vuille
Chapter1: Units, Trigonometry. And Vectors
Section: Chapter Questions
Problem 1CQ: Estimate the order of magnitude of the length, in meters, of each of the following; (a) a mouse, (b)...
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Hi, please explain why my answer is wrong (as the Wiley system stated) and also, I will appreciate if you include the correct answer. My solution looks like this, thank you
Your answer is partially correct. A student (m -63 kg) falls freely from rest and strikes the ground. During the collision with the ground, he comes to rest in a time of 0.04s. The average force exerted on him by the ground is +18000 N, where the upward direction is taken to be the positive direction. From what height did the student fall? Assume that the only force acting on him during the collision is that due to the ground. H- 6.7 eTextbook and Media GO Tutorial Hi, please explain why my answer is wrong (as the Wiley system stated) and also, I will appreciate if you include the correct answer. My solution looks like this, thank you: Chapter 7: 4, 11, 12, 19. Students may still have questions from chapter 6, so please kindly work with them if time permits. m SOLUTION We begin with the energy conservation principle m+mgh = mv+mgh (Equation 6.9b) applied to the student's fall to the ground. Falling from rest implies vo= 0 m/s, and the student's final velocity is the impact velocity: v=Vimpact . Thus, we have (1) 0+ mgh = 1 mv² 2 impact H = impact 2g + mgh or (2) Vo=Vimpact Substituting Equation (2) into Equation (1) yields m HI For the student's collision with the ground, the impulse-momentum theorem gives Fground Atmy-mv (Equation 7.4). The collision brings the student to rest, so we know that v=0 m/s, and Equation 7.4 becomes Fground At=-mv. Solving for the impact speed vo, we obtain Fground Af F 8(h-h)=2pct 2 impact ground Ar m 2g ² impact 2g II = or (Fground At) [(+18 000 N)(0.040 s)]² 2gm² 2(9.80 m/s²) (63 kg)² = 6.7 m
Transcribed Image Text:Your answer is partially correct. A student (m -63 kg) falls freely from rest and strikes the ground. During the collision with the ground, he comes to rest in a time of 0.04s. The average force exerted on him by the ground is +18000 N, where the upward direction is taken to be the positive direction. From what height did the student fall? Assume that the only force acting on him during the collision is that due to the ground. H- 6.7 eTextbook and Media GO Tutorial Hi, please explain why my answer is wrong (as the Wiley system stated) and also, I will appreciate if you include the correct answer. My solution looks like this, thank you: Chapter 7: 4, 11, 12, 19. Students may still have questions from chapter 6, so please kindly work with them if time permits. m SOLUTION We begin with the energy conservation principle m+mgh = mv+mgh (Equation 6.9b) applied to the student's fall to the ground. Falling from rest implies vo= 0 m/s, and the student's final velocity is the impact velocity: v=Vimpact . Thus, we have (1) 0+ mgh = 1 mv² 2 impact H = impact 2g + mgh or (2) Vo=Vimpact Substituting Equation (2) into Equation (1) yields m HI For the student's collision with the ground, the impulse-momentum theorem gives Fground Atmy-mv (Equation 7.4). The collision brings the student to rest, so we know that v=0 m/s, and Equation 7.4 becomes Fground At=-mv. Solving for the impact speed vo, we obtain Fground Af F 8(h-h)=2pct 2 impact ground Ar m 2g ² impact 2g II = or (Fground At) [(+18 000 N)(0.040 s)]² 2gm² 2(9.80 m/s²) (63 kg)² = 6.7 m
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