Free fall equations Sky diver Felix need some help please

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SUMMARY

The forum discussion focuses on calculating the free fall equations for skydiver Felix Baumgartner, specifically determining the time and distance to reach the speed of sound during his jump from 36,000 meters. The user initially neglected air resistance, leading to unrealistic results, and later incorporated a drag coefficient of 0.24, but still arrived at incorrect values of 2.7 x 10^56 seconds for time and 53 m/s for velocity. The discussion emphasizes the importance of accurately modeling air resistance to achieve realistic outcomes in free fall calculations.

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najat
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free fall equations ...Sky diver Felix :) need some help please

hi :)

OK..i tried to use the equations below to evaluate the distance or the time at which felix will reach the speed of sound ...but i failed ..i need some help:
here what i did:

first i neglect the air resistance , but the result was unreasonable so i used these:

jb13498311781.jpg


i assumed that :

m=weight of felix=70 kg
h=the height=36000 m
k=air resistance=0.24 >> i think "k" during the falling and by the time will be increased to 70

the result was :
t=2.7*10^56 sec :0
v=53 m/s :(

i don't know what i miss here ! so if someone will help me i will be so tankful ..
 
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hi ..
any body have any idea :(

by the way this is not for home work , it just a question pop up to me and i want to solve it , but it has been moved here :)
 


Just for info...

http://hypertextbook.com/facts/JianHuang.shtml

"An hour and thirty-one minutes after launch, my pressure altimeter halts at 103,300 feet. At ground control the radar altimeters also have stopped-on readings of 102,800 feet, the figure that we later agree upon as the more reliable. It is 7 o'clock in the morning, and I have reached float altitude …. Though my stabilization chute opens at 96,000 feet, I accelerate for 6,000 feet more before hitting a peak of 614 miles an hour, nine-tenths the speed of sound at my altitude."

Scroll down that page for the maths.
 

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