2007-10-30 02:35:46 · 7 answers · asked by Anonymous in Science & Mathematics ➔ Physics
Google it my A ss. Read the question. It has been over 20 years since I was in college. I'm lucky to be on the internet at all.
2007-10-30 02:40:56 · update #1
In a vacuum, v = at^2 (velocity = acceleration times time squared)
The acceleration above the earth is roughly 32 ft/sec^2, or 9.8 meters/sec^2
In the atmosphere, the terminal velocity is going to be affected by the density of the air and the shape of the falling item/body. Terminal velocity in Denver won't be the same as in LA for the same body due to the 5000 foot altitude difference and the consequential difference in air density.
In 1960(I think) a man jumped out the gondola of a balloon at 100,000 ft and hit almost 700 miles an hour before the denser atmosphere began slowing him. I've heard that parachutists jumping from 15000 feet can, by adjusting body position, attain speeds ranging from about 100 to about 200 miles per hour.
whoa....I just looked at the wikipedia article referenced by Misshicc..... and I see that that's probably closer to what you need rather than my straight velocity formula above.
2007-10-30 02:43:24 · answer #1 · answered by David Bowman 7 · 0⤊ 0⤋
This Site Might Help You.
RE:
20 years since physics...What is the formula to calculate terminal velocity?
2015-08-06 20:37:54 · answer #2 · answered by Cathleen 1 · 0⤊ 0⤋
The formula you're looking for is:
V = √((2mg)/(pAC))
where V is terminal velocity, m is mass of the falling object, g is the gravitaional acceleration (9.81 m/sec²), p is the density of the medium the object is falling through (1.269 kg/m³ for air), A is the object's cross sectional area, and C is the object's drag coefficient. You should be able to figure this out, the only problem being of that finding the drag coefficient For a man in an upright position, it's roughly 1.0, for a man in prone position, about 1.3, and for a rough sphere, about 0.4, while a smooth sphere is about 0.1.
2007-10-30 03:08:48 · answer #3 · answered by Scythian1950 7 · 1⤊ 0⤋
terminal velocity = limit velocity
mV^2 /2 = m g h
V^2 = 2 g h
V(terminal) = squareroot (2 gh )
.
2007-10-30 02:43:13 · answer #4 · answered by Anonymous · 0⤊ 1⤋
http://en.wikipedia.org/wiki/Terminal_velocity
formula is listed on article
2007-10-30 02:39:04 · answer #5 · answered by misshiccups 3 · 0⤊ 0⤋
http://www.ajdesigner.com/phpstokeslaw/stokes_law_terminal_velocity.php
Go there..Hop3 that H3lpz!!
2007-10-30 02:44:50 · answer #6 · answered by Courtney 3 · 0⤊ 0⤋
google it!
: )
2007-10-30 02:38:19 · answer #7 · answered by Anonymous · 0⤊ 1⤋
m/s?
2007-10-30 02:38:37 · answer #8 · answered by Anonymous · 0⤊ 1⤋