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.