Help for Screenwriter on Stan Lee project

[Neil B]

A man weighs 72 kg. he jumps from a tower 287m high. No windspeed. What appoximate speed wil he hit the ground in Km/hour?

Please help.

 

[phinds]

Do you reckon his weight matters?

 

[Dr. Courtney]

Does he fall mostly head first, feet first, front first, or back first?

 

[phinds]

Does he call mostly head first, feet first, front first, or back first?

He usually calls head first, but when he's had a lot of beans, sometimes it's back first.

 

[Dr. Courtney]

I don't think one can ignore air resistance on this one. It seems likely to be in the region before terminal velocity but after air resistance has become significant. This means the best approach is to estimate frontal area and coefficient of drag and numerically integrate the diff eq. Mass matters here.

 

[Neil B]

Assume he falls head first. that's what we've already filmed. Like a swallow dive. The actors about 5 ft 8. I'd say 42 inch chest. If that helps?

 

[Bandersnatch]

There's this calc here:
http://keisan.casio.com/exec/system/1231475371
The air resistance coefficient they use is 0.24 kg/m, which they say is typical for skydiving. They don't say which position it is, though.

I tried calculating the coefficient from $F=\frac{1}{2}qC_dAV^2$, where I took q to be 1,2 kg/m^3, C_d for an upright man to be 1 (from here: https://en.wikipedia.org/wiki/Drag_coefficient) and the area equal to 0,12 m^2 (42" chest is about the same cross-section as a circle with 0.16m radius, plus a bit extra for arms, say r=0.2m). It nets an air resistance coefficient of about 0,07 which I think would be about right if the one in the calculator is for a horizontal falling position.

Plugging that into the calc, you get 360 236 km/h for an upright position, and 200 180 km/h for a fall in a horizontal position. These numbers are consistent with (i.e. lower than) terminal velocities in skydiving as found e.g. here:
http://hypertextbook.com/facts/JianHuang.shtml

(edit: wrong numbers - forgot to change height)

 

[Dr. Courtney]

One of my basic principles is that the answer to a hard problem is often between the answers to two much easier problems. In this case, the impact velocity will be lower than that obtained ignoring air drag.

So first, we compute the impact velocity ignoring air drag.

Using conservation of energy, the kinetic energy at the bottom (0.5 m v^2) will equal the potential energy at the top (mgh).

The mass cancels out so solving for v yields v = sqrt(2*g*h) = sqrt(2*9.8 m/s/s * 287 m) = 75 m/s.

Converting to km/hr yields v = 270 km/hr.

 

[Bandersnatch]

higher than the terminal velocity.

Why would that ever be the case?

 

[Dr. Courtney]

Edited a correction. Good catch.

 

[jackwhirl]

How many seconds before he reaches the ground? Will the screenwriter have the man do improbable things on the way down?

 

[Bandersnatch]

How many seconds before he reaches the ground? Will the screenwriter have the man do improbable things on the way down?

About 8 seconds.

 

[Neil B]

Amazing. Thank you...

 

[meBigGuy]

distance = 1/2 (a) (t^2)
t = time
a = acceleration of gravity = 9.81 m/s
distance = 287m
t = sqrt (287/(9.81/2)) =7.64 seconds in a vacuum

velocity = a * t = 7.64 * 9.81 = 75 meters/sec ( terminal velocity for a body in air is 56m/sec and takes 12 seconds, so this is wrong for a real fall in air)

The full solution with air resistance is in here.
https://en.wikipedia.org/wiki/Free_fall

This takes you through it also
http://www.wikihow.com/Calculate-Terminal-Velocity

Turns out the mass matters when you consider air resistance.
I'm not sure what numbers to use for drag coefficient since it depends on the orientation of the body.

 

[meBigGuy]

This has some pretty good graphs
http://arachnoid.com/sage/terminal_velocity.html

Looks like 287 meter fall in air will take ~9 seconds and reach ~50m/sec