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Harrisza Parchute Lab Report
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Harrisza Parchute Lab Report

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  • 1. Zach Harris Mrs. Shea Period 2 24 November 2008 Partners: Jake Boykin, Craig Cilley, David Custer Design a Parachute Recorded Drop Times: Trial 1 Trial 2 3.81s 2.98s The experiment was designed to test a parachute using different methods of creation and ideas to come up with one that would reach terminal velocity faster. By reaching terminal velocity, the force of gravity downward equals the drag force going upwards. As the parachute accelerates downward, the drag force should increase. Enventually during the fall the drag force should equal the parachute’s weight meaning that the acceleration is zero. That would be the terminal velocity because the object is now in constant velocity. Reaching a terminal velocity fast was good in this experiment because the parachute would then take longer to reach the ground. Since it would not be accelerating, the time would be longer. If the time was long, then the object reached terminal velocity faster than if the time was short. Using Logger Pro the group was able to determine when the parachute reached terminal velocity. The focus was only on the y-axis portion. Using linear fit the group was able to determine where terminal velocity had been reached. The slope of the line equals meters per second and a straight slope equals the constant velocity. This is where the terminal velocity. Newton’s second law indicates that when the mass is zero, then the acceleration will be zero. Thus, the object will be in constant velocity and will stay that way due to Newton’s first law which states that an object in constant motion will remain in constant motion until a net force acts upon it. The net force acting on the parachute is zero when it is in terminal velocity. When calculating b, the net force equals the air resistance minus the force of gravity. When setting up the problem, the force of gravity will equal the air resistance. Then the fg was changed to mg and the air resitance is changed to bv. When calculating b the mass(in kilograms) multiplied by 9.81 (gravitational force), then divided by the terminal velocity will equal b. b=mg/v b=(.66)(9.81)/12.6
  • 2. b=.514 Other Group Data: Group Trial 1 (s) Trial 2 (s) b Erika 2.82 3.53 .179 Bethany 6.89 7.31 .657 Morgan 6.42 4.90 .364 Andrey 2.42 2.23 .546 Angela 4.86 6.83 .496 The drop times show somewhat of a correlation with b through an indirect way. If the times were long, the terminal velocity was reached at a faster time which means that it is small. If the terminal velocity is small, then the b should be a larger number. If the experiment could be done again, the parachute would have more surface area with the trash bags and made with less mass. If there was no wind factor, the new parachute should reach terminal velocity faster and thus take longer to touch the ground. The new chute would be made wider to catch more of the drag force going up. I have neither given nor received help on this assigment. Zach Harris.