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.

Tikkabhuna · 9 May 2011 at 13:59

I would work it out the same way as you. Ignoring wind resistance that would be the easiest way.

Vonhelmet · 9 May 2011 at 13:59

If you're ignoring any forces besides mavity, then yes - you can do it with v=u+at.

If you want to model air resistance as well, then you have to use differential equations, as the air resistance is proportional to the velocity, so the faster you go the more force acts against you, until eventually you reach terminal velocity when the forces are balanced.

Perhaps that is what your friend is thinking of?

Or perhaps he's thinking of how you work out the distance travelled by the time you hit a given speed, which would be calculated from v^2=u^2+2as.

Tokenbrit · 9 May 2011 at 14:05

You are correct.

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!

Alex74 · 9 May 2011 at 14:06

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.

thepeganator · 9 May 2011 at 14:07

You are correct.

But wind resistance is the thing, then you start getting complicated(er, it's not that bad tbh...).

And doing Maths at Uni, so tell him to stfu.

Alex74 · 9 May 2011 at 14:08

thepeganator said:
...doing Maths at Uni...

*hi5*

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?

thepeganator · 9 May 2011 at 14:18

Alex74 said:
*hi5*

*5's*

Vonhelmet · 9 May 2011 at 14:31

Amp34 said:
Is v^2=u^2+2as really a quadratic?

Assuming u is nil, then yes - in theory it is. However, in practise, you can't have a negative solution to it, so it isn't entirely relevant or helpful to think of it in those terms.

Duff-Man · 9 May 2011 at 14:41

In a vacuum then yes, the skydiver would fall as a particle (uniform linear acceleration, which leads to the equations you quote).

In practice, you will see an opposing force proportional to the square of the velocity (due to air resistance). In order to solve this problem you will need to set out the differential equation with the gravitational force (constant) and the opposing air resistance ( ~k*v^2). When you solve it you will see that the skydiver rapidly approaches a maximum fall velocity, where the force of mavity is balanced by the air resistance. This is called the terminal velocity.

The terminal velocity is very difficult to compute since it will be radically different for different shapes (it's a property of how the airflow interacts with the shape that's falling). However, if you can measure or estimate the terminal velocity, you can get an estimate for the parameter "k" (which describes mathematically the "difficult" parts relating to fluid-structure interaction). Once you have k, you can plug it back into your original solution and can estimate how far the skydiver has fallen at any given time.

In short, to answer the question you're asking you will need to know:

a) The terminal velocity for the skydiver
b) How to solve a linear second order ODE

Duff-Man · 9 May 2011 at 14:51

Amp34 said:
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).

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.

wikipedia said:
a speed of 50% of terminal velocity is reached after only about 3 seconds, while it takes 8 seconds to reach 90%, 15 seconds to reach 99% and so on

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.

sbweightman · 9 May 2011 at 14:54

I gave up on maths when there were more letters than numbers in an equation.
But simple equation does look right.

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!

The Dread Pirate · 9 May 2011 at 15:58

That would be to calculate his velocity without wind resistance and probability of hitting somthing(Bird, Plane, Superman etc..) if you just want a rough estimate you're on the right track if you want more in depth you've got to take other things into account.

Morbius · 9 May 2011 at 16:52

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.

Vonhelmet · 9 May 2011 at 17:03

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.

Air resistance will be a far greater factor than the variation in mavity over the distances involved.

The acceleration felt by a body of 70Kg at 5km above the earth (estimated skydiving height) is about 9.78m/s^2.

The acceleration felt by a body of 70Kg at the earth's surface is about 9.81m/s^2.

Lightnix · 9 May 2011 at 17:21

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

Morbius · 9 May 2011 at 17:23

vonhelmet said:
Air resistance will be a far greater factor than the variation in mavity over the distances involved.

Absolutely, but the original post said no air resistance.