# Homework Help: Soving an eqn with exponential & linear parts

1. Jan 7, 2008

### Ikeness

1. The problem statement, all variables and given/known data
Known Constants
A, B and C

Unknown (we are solving for)
t

e is the exponential as in ln(e) = 1

2. Relevant equations
$$e^{t*A} + t*B + C = 0$$
(we want to solve for t)

3. The attempt at a solution
This is not for school or anything.
This is actually part of an attempt at the solution of another problem. Reaching this equation I had to stop. I do not remember seeing something like this in Calculus or Differential Equations, so I need help.

My overall purpose is to find the initial velocity necessary to launch a particle to a specified location. This particle is subject to gravity and air drag. If any of you happen to have the solution to this (or something simular) on hand, that information would be worlds better than just getting me past this one step.

I have my progress somewhat documented in this ugly text file.
If you care to read on, feel free. But it is not necessary.

Last edited: Jan 7, 2008
2. Jan 7, 2008

### nicksauce

Equations with exponential and linear parts cannot be solved analytically in general. At best you can make a second order taylor approximation and solve the resulting quadratic.

3. Jan 7, 2008

### blochwave

You usually write the resisting force as F=kmv, not kv

so the y direction equation of motion is (' is dot! Denoting time derivative) my''=-kmy'-mg, the resistive force always opposes the velocity, hence its negative, and the gravitational force is down

dv/(kv+g)=-dt

1/k*ln(kv+g)=-t+C

kv+g=e^(-kt+kc)

knowing initial y velocity we assume is Vyi at t=0

Vy=-g/k+[k*Vyi+g]*e^(-kt)/k

Vy=dy/dt, so you can then integrate again to find y(t)

The x equation is similar but mx''=-kmx' and that equation is a little easier

That's for you to compare to your own work(which I see at a glance is different, but perhaps not woefully so), it's straight out of my classical mechanics textbook. Also mentioned in my classical mechanics textbook is that attempting to solve for things like time of flight and stuff can't be done analytically because the equations become transcedental, and unless you know much perturbation theory or have powerful numerical solving tools, you're done, and I bet that's the issue you're having right now

An easier problem to solve is when the resistive force is proportion to v SQUARED. Much cleaner solution and can be done analytically, and, like a linear dependence on v, is realistic for certain circumstances

4. Jan 7, 2008

### Ikeness

Yes. Thanks. I was carrying around an extra m. I also really screwed up on that exponential step as I have noted above. But asside from that, our problems steps are very simular.

Running through with v SQUARED resistive force I stopped after the first integration step to make sure that I am doing it right. Please let me know.
Thanks.
-Joel

5. Jan 7, 2008

### blochwave

Sorry you typed all that out but

dv/(-kv^2 - g) = dt
dv/(kv^2 - g) = -dt

Look hard, what went wrong there?

6. Jan 7, 2008