Speed of a person jumping off a building?

[Sylvias]

Alright, I need this information not for science homework, but for writing a short story. You see, I am very particular about the tiny details, and in the story there is going to be a superhuman jumping off a building.

I want to know the speed of such a jump to dramatise the description, but I have no idea how to calculate it. With the limited knowledge of v=d/t and f=ma and some sketchy understanding of terminal velocity (from the simple to understand wikipedia), I doubt I ever will find the solution without help.

Suppose the height of the building to be 410 metres, the weight of the person jumping to be of an average adult male (I have no idea of this), the jump to be roughly straight down, and the wind to be calm.

I have no idea what drag coefficient is, so I can't say, and as for the projected area... just assume that he isn't moving very much? (And is the projected area measured at the feet or the body width?)

Hopefully I have provided enough information to calculate the speed. If not, just make up the numbers, and provide the workings, and if possible, explantions. (Extra, comprehendable, knowledge is always fun and useful)

Thank you. :D

 

[jeffreydk]

The formula for terminal velocity would be,

$$V_t=\sqrt{\frac{2mg}{\rho A C_d}}$$

where $m$ is the mass of your jumper in kilograms $g=9.81$ meters per second squared, and there are more specific things involved in the formula like the density of air ($\rho$) the drag coefficient ($C_d$) and the area of the falling object that is responsible for the drag. You can see the terminal velocity changes as the environment changes or the body position changes (changing the body area and the density of air), but typically I would say terminal velocity is about 120 miles per hour and you reach it after about 4 and a half seconds of falling.

 

[Nick89]

There is always the question if he will even reach terminal velocity. You don't fall at terminal velocity as soon as you jump, terminal velocity is just the velocity where you stop accelerating (because the pull of gravity pulling you down is exactly matched by the force of air resistance pushing you back up). But I bet you knew that already ;)

 

[jeffreydk]

Yea, I estimated about 4 1/2 seconds for terminal velocity.

 

[Nick89]

I was replying to the original poster actually but ok :)

I just noticed he told us the building is 410 meters high. I'm pretty sure you reach terminal velocity from that height (although I didn't calculate it) so that won't be much of a problem.

 

[jeffreydk]

oh whoops hah sorry.

yea i agree, I am pretty sure they'd reach it.

 

[Sylvias]

So according to your workings...

Terminal velocity of a falling person = 196 km per hour
= 54m/s
Reached after 4.5s
I.E. After falling 245m
Fell for roughly 3s after hitting terminal
Total time = 7.5s//

Hopefully I didn't get that simple Maths wrong. But there is one thing I do not understand, and that is how you came to the 4s fall to terminal velocity. I guess that acceleration (or is it velocity? I don't really know. Could you explain this as well?) has to be into account, but I have any clue how beyond that.

Sorry for being troublesome, and advance thanks if you do answer :D

 

[PhanthomJay]

So according to your workings...

Terminal velocity of a falling person = 196 km per hour
= 54m/s
Reached after 4.5s
I.E. After falling 245m
Fell for roughly 3s after hitting terminal
Total time = 7.5s//

Hopefully I didn't get that simple Maths wrong. But there is one thing I do not understand, and that is how you came to the 4s fall to terminal velocity. I guess that acceleration (or is it velocity? I don't really know. Could you explain this as well?) has to be into account, but I have any clue how beyond that.

Sorry for being troublesome, and advance thanks if you do answer :D

The math is not that simple, since acceleration is not constant and non linear, as you can see from this site:

http://hyperphysics.phy-astr.gsu.edu/hbase/mechanics/fallq.html

As noted in prior posts, the terminal velocity (which is approached, but never reached) of a skydiver is somewhere around 120mph (55m/s), but it depends on whether the diver is falling head or feet down, curled up in a ball, or belly up, because this changes his/her area and shape factor exposed to the rushing air (belly up leads to slower terminal velocity, maybe in the order of 100 mph or so). In any case, if the person were in free fall without air resistance, it would take about 6 seconds to reach 120mph, so that since the acceleration with air drag is always less than 'g' (it varies from g down to near 0), I'm guessing it takes about 8 seconds or so to reach 90 percent of terminal velocity, reached at a height of some 300 meters below the top of the building, and about 11 or 12 seconds to hit the ground at 410 meters. By comparison, without air drag, the person would hit the ground in about 9 seconds, at a speed of about 200 mph . These are approximate numbers. actual calculations are a bit complex, unles you like the hyperbolic functions.

 

[DaveC426913]

If you were to ignore terminal velocity, I get a final velocity of 63m/s, or about 140mph.

Since that's more than terminal velocity, it follows that we should have taken it into account.