Maths question - Acceleration/freefall

Amp34 · 9 May 2011 at 13:56

A friend and I have been debating this recently and I want to check if I'm right and if not why...

The question is based on a freefalling object, say a skydiver jumping out of a helicopter and the aim is to work out how fast they would be falling after a certain period of time, or a certain distance.

Now my assumption was to use UVATS equations, more specifically the simplest:

v=u+at

v=final velocity
u=initial velocity
a=acceleration
t=time

With final velocity (v) being what I want to work out after a set period of time (t). u would be 0 as we are starting from time "0" and a would be a constant rate, specifically 9.81m/s^2.

So all in all you'd hit around 50m/s (176Kmh) after 5 seconds of falling.

However my friend swears it isn't that simple and you need to use quadratics. So the question is - who is right?

**calculations use the standard rules of no wind resistance etc and using only vertical moments/force direction.

Amp34 · 9 May 2011 at 14:06

One of my thoughts when typing the OP was whether he was thinking of air resistance interaction but I specifically pointed out I wasn't modelling it due to it's insignificance at this sort of time (would be max 5 seconds we were talking about). Do you know what differential equation you would need to use i'll try and work it out and see what the difference is (it's been a few years now since my A-Level maths and Physics...).

I don't think he was thinking about any of the uvats equations as I did mention them all to him. Either way that's certainly not a quadratic (AFAIK?).

EDIT: Bejesus! This thread is already in the google results!

Amp34 · 9 May 2011 at 14:14

Alex74 said:
You'd need quadratics if you're working with distance as vonhelmet says . But assuming constant acceleration then velocity with no air resistance will be a linear function of time.

Is v^2=u^2+2as really a quadratic?

Amp34 · 9 May 2011 at 14:58

Duff-Man said:
Well, as you point out five seconds of freefall in a vacuum would lead to a speed approaching 50m/s, however the terminal velocity of a freefaller is only around 55 m/s according to wikipedia. So yes, air resistance becomes very important well before 5s of freefall.

That should give you a ballpark estimate of how fast you'll be going after 5s. Something in the order of 75% your terminal velocity, so around 40m/s in the "belly out" position.

I was thinking more like 2 seconds and took something from another wiki page that suggested it would be quite innacurate after 5 seconds, didn't realise it was that innacurate.

I was also thinking more of a pencil dive when I was thinking this through (aka a lot less air resistance), but in actual fact the belly out position would make more sense!

Amp34 · 9 May 2011 at 18:24

Morbius said:
Assuming that mavity is constant, then v=u+at is fine.

mavity isn't like that in real life though. It varies with the inverse square of the distance between the two objects.

Lightnix said:
Guys you forgot to factor in that the falling object could be non-spherical and turning.

Haha, I was waiting for this sort of thing to come up...

I know mavity isn't constant, but for basic calcuations the variation is so minor it's not really going to make a difference, as von said.

And yep, just class it as a point mass I'd suggest.

This discussion all stemmed from a (what I presumed) basic back of a napkin calculation and sort of ballooned...