# Modeling a falling body

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1. Feb 28, 2016

### Anthony Salls

1. The problem statement, all variables and given/known data
The speed of a falling body might be based on the observation that the velocity of a falling object seems to increase the further it has fallen. Model the hypothesis "The speed of a falling object is proportional to the distance it has fallen" as a differential equation initial value problem. By analyzing the predictions of your model, explain why this "law of gravity" could not be correct.

2. Relevant equations
N/A

3. The attempt at a solution
So, obviously, v(t)=v0+k*d(t), where k is the proportionality constant.
d(t) = (v0+v(t)/2)*t
But plugging in d(t) and solving yields -(2*v0 + k v[0])/(-2 + 1*k) which if v0 = 0 is always 0 and there are restrictions on k due to the numerator. Also, it's not a differential equation and I'm not sure how to get one. I tried setting dv/dt = a and solving that way but it yields a nonsensical equation for v(t).

So my question is, how do I set up the model?

2. Feb 28, 2016

### Staff: Mentor

How did you get this?

v0=0 is a problematic starting condition. I would exclude it here and consider other cases. You don't need it for the differential equation anyway.

3. Feb 28, 2016

### LCKurtz

If you call the position at time $t$ by $s(t)$, then $v(t) = s'(t)$. Start by expressing "The speed of a falling object is proportional to the distance it has fallen" as an equation in $s(t)$.

4. Feb 28, 2016

### Anthony Salls

It's the average velocity which would have units of distance/ time multiplied by time so it would be the distance as a function of time.

So s(t) = c(v(t)-v(0)) (rearrangement of v(t) = v(0) + k*s(t) with 1/k = c )
Alright, so if I'm understanding this correctly ds(t)/dt = v(t), and therefore ∫ds =∫v(t)dt so v(t) = v(0)+cinte^(c*t) then log[v(t)] = log[v(0)]+c*t and therefore v(t) = v(0) + e^(c*t)
and the reason this is a poor model is because the acceleration is not constant because it is tied to e^t and therefore overestimates the acceleration of the object.

5. Feb 28, 2016

### LCKurtz

That is not a correct rendition of the above quote in terms of $s$. Try again.

6. Feb 28, 2016

### Anthony Salls

I don't understand why it isn't. Forget the initial velocity? But that doesn't work either. ||v(t)|| ∝ s(t)-s(0),
so v(t) = k*(s(t)-s(0)). -> v(t)/k +s(0) = s(t), s'(t) = v(t). -> d/dt(v(t)/k + s(0)) = v(t), v(t) -> 0

EDIT: okay so I took the integral of v(t) and set it equal to v(t)/k +s(0) and solved for v(t), which = k*s(0)/(-1+k*t) but this has weird things going on like the initial position cannot equal 0 or else the velocity is 0, and k*t cannot equal 1

Last edited: Feb 28, 2016
7. Feb 28, 2016

### LCKurtz

Remember $v(t) = s'(t)$ so the equation I have highlighted in red can be written $s'(t) = k(s(t)-s(0))$. Solve that DE for $s(t)$ and you will also know $v(t)$. Then you will have the correct solution to analyze.

That makes no sense to me. How can you take the integral of v(t) when you don't know what it is?

Last edited: Feb 29, 2016
8. Mar 1, 2016

### Staff: Mentor

It is not the average velocity, in particular not if you don't even know the acceleration profile. There is an easy formula for a constant acceleration (this formula is different from what you wrote), but you do not have a constant acceleration.