I realize there are variables to take into consideration, I would like to find out the minimum distance in feet, it would take an average sized man in a feet first dive...
2006-08-29 15:51:29 · 11 answers · asked by Anonymous in Science & Mathematics ➔ Physics
mass is completelly irrelevant for falling bodies. this was prooved hundreds of years ago.. amazing to see a lot of people still have to catch up with it.
You have to assume a terminal velocity.. which is really dependent on the posture of the diver, altitude and so on.
Let's assume 100 m/s.
Gravity will accelerate a body at 9.8 m/s^2 so to get from 0 to 100 m/s u need 100/9.8=10.20 seconds.
You can consider it as uniform accelerate motion from 0 velocity.. thus use the space formula:
S=1/2*A*T^2
with A=9.8 m/s^2
and T=10.20 s
resulting in around 1019m.
Repeat the calculation for a given terminal speed.
2006-08-29 23:29:05 · answer #1 · answered by kunosayu 2 · 0⤊ 0⤋
Depends on whether they arch and spread their limbs, or "dive" headfirst with arms tight to the body and toes pointed. An experienced skydiver can reach about 140 m/s, the low end for a "falling human" is about 50 m/s. A "typical" skydiver is at about 60 m/s. It takes 10-12 seconds to reach those velocities.
It also depends on the altitude. In the thin air above 100,000 ft (from a balloon gondola), the record is 274 m/s.
Aloha
2006-08-29 15:55:46 · answer #2 · answered by Anonymous · 0⤊ 0⤋
There is a stellar article on this at http://en.wikipedia.org/wiki/Terminal_velocity.
Basically, you need to approximate the mass, cross-sectional area, and coefficient of drag. Please note, the terminal velocity is not a constant number because the atmospheric density changes with altitude due to the weight of the air. Atmospheric density increases as you fall which lowers your terminal velocity about 1% every 160m. Gravitational acceleration changes as well, but not to the extent of atmospheric density.
For the average man in a feet first dive, density equals 1.0g/cm^3, mass is 200lb or 890N, the air density is roughly equal to 1.2 kg/m^3 at 20C and 1atm, the cross sectional area for a human standing is about .19 m^2, and the coefficient of drag can be estimated at 1 per historical data viewed online. Using Vt = sqrt(2mg/rho*A*Cd) = 88.4 m/s or 197.7 MPH. Of course, with your arms open, you increase your cross-sectional area and decrease your terminal velocity.
2006-08-29 16:57:04 · answer #3 · answered by mcfallsg 1 · 0⤊ 0⤋
Impossible to answer.
Terminal velocity isn't a set value. When gravity is pulling you just as much as the atmosphere is pushing you back, you've reached terminal velocity.
Terminal velocity depends on how high you've jumped from, and thus the time to reach it varies. You can break the sound barrier a couple times over if you jump from high enough.
2006-08-29 17:36:27 · answer #4 · answered by Anonymous · 0⤊ 0⤋
Objects fall at a constant rate of 32 ft per second, per second regardless of mass. Their terminal velocity will be somewhere between 100MPH (spread out) and 200MPH (falling stiff/straight).
I haven't calculated the feet, but it's a few hundred.
2006-08-29 16:07:04 · answer #5 · answered by Anonymous · 0⤊ 1⤋
Around 125 feet...
2006-08-29 16:00:12 · answer #6 · answered by 345Grasshopper 5 · 0⤊ 0⤋
i dont think a human can reach terminal velocity....
I know cats are and other animals can at a relatively low acceleration rate.... but humans just keep plunging...
2006-08-29 16:00:01 · answer #7 · answered by Anonymous · 0⤊ 2⤋
Well, I think that if he doesn't have a parachute, whatever velocity he reaches, it will be terminal!
2006-08-29 19:36:23 · answer #8 · answered by walkingteapot 1 · 0⤊ 0⤋
Everyone except mcfallsg any maybe enufwork has absolutely no idea what they're talking about.
2006-08-29 18:25:35 · answer #9 · answered by Mikey C 1 · 0⤊ 0⤋
impossible to figure without knowing the mass.
2006-08-29 15:55:40 · answer #10 · answered by lowrider 4 · 0⤊ 0⤋