Assume a ball with a parachute attached to it (which has a mass of .12499 KILOGRAMS) is dropped from a height of 6.3599

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answerhappygod
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Assume a ball with a parachute attached to it (which has a mass of .12499 KILOGRAMS) is dropped from a height of 6.3599

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Assume A Ball With A Parachute Attached To It Which Has A Mass Of 12499 Kilograms Is Dropped From A Height Of 6 3599 1
Assume A Ball With A Parachute Attached To It Which Has A Mass Of 12499 Kilograms Is Dropped From A Height Of 6 3599 1 (51.13 KiB) Viewed 49 times
Assume A Ball With A Parachute Attached To It Which Has A Mass Of 12499 Kilograms Is Dropped From A Height Of 6 3599 2
Assume A Ball With A Parachute Attached To It Which Has A Mass Of 12499 Kilograms Is Dropped From A Height Of 6 3599 2 (36.98 KiB) Viewed 49 times
Assume A Ball With A Parachute Attached To It Which Has A Mass Of 12499 Kilograms Is Dropped From A Height Of 6 3599 3
Assume A Ball With A Parachute Attached To It Which Has A Mass Of 12499 Kilograms Is Dropped From A Height Of 6 3599 3 (55.99 KiB) Viewed 49 times
Assume a ball with a parachute attached to it (which has a mass of .12499 KILOGRAMS) is dropped from a height of 6.3599 meters. The following tables and graphs illustrate its velocity vs time // position vs time motion. The ball starts at a velocity of -1.14m/s, however, was technically at rest. The data is recorded the moment the ball is released. Assume the motion is only given in the y direction, therefore the X motion is nonexisting. Given the information compute the following: A.) Draw and label in detail a quantitatively competent physics energy bar chart (with numerals) for the exact moment the ball is released, its halfway journey, the moment immediately before it approaches the terminal, and when it is finally at rest after its journey. B.) Explain your charts using intuition and evidence (the evidence needs to be consistent with your graphical observations) and detail concerning the energy that is being stored. Be sure to have your calculations labeled clearly. BE SURE TO CHECK ALL THE DATA BEFORE COMPUTING WITH THE LABS COLLECTED LAST WEEK.

* When released the ball has a t=0 seconds, position of y = 6.359 meters. A velocity of -1.14m/s. * At it's halfway journey, the ball has t=.82 seconds, Y = 3.6 meters. velocity of 4.47m/s. * The moment before it hits the surface, t= 1052 seconds, Y = .0 64 meters, velocity of 0.21 m/s s A period (short) of impact, t= 1.66, Y=0 meters, velocity of .. 18 m/s. s

Position (meters) Mr. Meyer Table $ Graph Legend : AP Physks 10 Phase I3rd phase Position vs time graph 12nd Phose 4th Phase [0-185] [6.359-617] Meters 0.24-.82 seconds 16.0-3.7 meters] 1.84-1.43 seconds) (3.6-47 meters] 11.45-1.66 Seardit] 30-4-0 meters TIME SECONDS . fhose Present Time interval Slorking relateerd velocity at end of time TABLE Phase I 0-18 seconde o seceras -1.14 M/S at -2.35m/s at 18 Secords The s . is whats at the Phase2 C24-182] -3m/s at 24.5 - 4. uzmys at 9825 [ begining of time Phase 3 [84-1.43) | -Yalmis at -4.38m/s at 1.435 Interval. -4.86m/s at 18 m/s at 1.6s. .84.5 Phase 4 [1,45 1,65 1,455
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