Projectile and time of travel

1. Jul 12, 2008

loonychune

1. The problem statement, all variables and given/known data

Show that, in the limit of large damping, the time of flight of a projectile (the projectile is fired from level ground) is approximately,

$$t \approx \left(\frac{w}{g} + \frac{1}{\gamma}\right)\left(1-e^{-1-\gamma w/g}\right)$$

2. Relevant equations

The equation of motion for the projectile is given by,
$$m\ddot{r} = -\gamma mr -mg$$
so that,
$$\ddot{z} = -\gamma \dot{z} - g$$
The solution to this equation is,
$$z = \left(\frac{w}{g} + \frac{g}{{\gamma}^2}\right)\left(1-e^{-\gamma t}\right) - \frac{gt}{\gamma}$$

3. The attempt at a solution
I arrived at the answer by saying that the,
$$e^{-\gamma t}$$
term vanishes for large gamma, so setting z = 0 i have,
$$t = \frac{w}{g} + \frac{1}{\gamma}$$
which you can then substitute back in to the exponetial term that vanished for t --- this gives the answer but the method seems terrible to me, i just saw it.

I thought of expanding the exponential about t = 0 but i'm not getting there.

I think i'm missing something obvious. It's looking at making approximations to solutions is this question and i think i'd find it pretty useful once i actually get throught it, so i'd appreciate your help.

Thank you.

2. Jul 13, 2008

dynamicsolo

I feel the need to ask whether w is the launch velocity. Units of velocity seems to be required to make the ratios dimensionally correct, but your problem statement doesn't say what it is. (If one uses v(0) = w as an initial condition, though, w does turn up in the right spots...)

After working out the differential equation myself, I'm finding the solution for the position to be

$$z = \left(\frac{w}{\gamma} + \frac{g}{{\gamma}^2}\right)\left(1-e^{-\gamma t}\right) - \frac{gt}{\gamma}$$

Again, the first term makes sense dimensionally with gamma there, but not with g...

There is another issue as well. There are two separate regimes to consider. We have

$$\ddot{z} = -\gamma \dot{z} - g$$

on the way up, but

$$\ddot{z} = \gamma \dot{z} - g$$

on the way down. Is the question asking for the total flight time?

Last edited: Jul 13, 2008
3. Jul 14, 2008

loonychune

Sorry for the lapse in time before this here reply...

Yes, v(0) = w is the initial velocity.
Your solution is where I got to.

The question is indeed asking for the total flight time. This is why i'm looking to substitute for z = 0 (it lands on level ground, at z=0), in which case one solution for t (given t = 0 initially) ought to yield the time of flight.

I looked to expand the,

$$e^{-\gamma t}$$

term so that, after setting z = 0, i might then group terms in t and then rearrange a few things.

The answer is given in part [1] of my initial post only I haven't been able to arrive at it.

Your help is much appreciated! Thank you...

Damian

4. Jul 14, 2008

dynamicsolo

It isn't quite that simple. The effect of drag (the "damping" in this problem) makes the trajectory asymmetrical in time. This must be treated in two parts because the forces on the projectile act differently on the way down from the way they do on the way up. (The velocity will fall to zero on the way up, but will only reach "terminal velocity" on the way down.)

For the path upward, you already have the velocity function, which can be solved for the time at which it reaches zero. That gives you the time spent traveling upwards and lets you find the maximum altitude.

You will then need to find the position function for the trip back down, which must satisfy

$$\ddot{z} = \gamma \dot{z} - g$$

Since we know the maximum altitude, we can work out an expression for position which would be set to zero to find the trip time down. I suspect this is where we'll really need that "large damping" approximation.

5. Jul 17, 2008

dynamicsolo

On working with this a bit more and graphing it, I find that it is sufficient to use the single equation

$$\ddot{z} = -\gamma \dot{z} - g$$

based on upward being positive, for the entire trajectory. The steady-state velocity is

$$v_{term} = -g/\gamma$$

which is the terminal velocity on the fall back (negative means "downward").

If you plot the position equation

$$z = \left(\frac{w}{\gamma} + \frac{g}{{\gamma}^2}\right)\left(1-e^{-\gamma t}\right) - \frac{gt}{\gamma}$$

with "large damping" (for instance, setting w = 300 m/sec and gamma = 1), you find that the projectile spends relatively little of the total flight time on the ascent and most of the descent is covered at the terminal velocity. This is why you find that you could just divide the coefficient $$\left(\frac{w}{\gamma} + \frac{g}{{\gamma}^2}\right)$$ by $$g/\gamma$$ to get the coefficient $$\left(\frac{w}{g} + \frac{1}{\gamma}\right)$$ in the equation for flight time.

Otherwise, what I described for the method should be all right. Setting the projectile velocity

$$v = [ (w + \frac{g}{\gamma}) · e^{-\gamma t} ] - \frac{g}{\gamma}$$

equal to zero allows us to find the time at which the peak of the trajectory is reached,

$$T = \frac{1}{\gamma} ln(1 + \frac{\gamma w}{g})$$

We then use this to find the maximum altitude for the projectile from the position equation

$$z_{max} = \left(\frac{w}{\gamma} + \frac{g}{{\gamma}^2}\right)\left(1-e^{-\gamma T}\right) - \frac{gT}{\gamma}$$

In the case of large drag, the first term greatly dominates the second, since w is also rather large, reducing this to

$$z_{max} \approx \left(\frac{w}{\gamma} + \frac{g}{{\gamma}^2}\right)\left(1-e^{-\gamma T}\right)$$ .

and the total flight time to

$$\tau \approx \left(\frac{w}{g} + \frac{1}{\gamma}}\right)\left(1-e^{-\gamma T}\right)$$ .

The one part I don't see is how the originator of the problem came up with the exponent for e . You can't use an approximation for the logarithm in T because $$\frac{\gamma w}{g}$$ is much larger than 1 . The exponential term should be

$$e^{-\gamma T} = e^{-ln(1 + \frac{\gamma w}{g})} = \frac{1}{1 + \frac{\gamma w}{g} } = \frac{g}{g + \gamma w}$$

So I'm rather unclear on just what approximation they are using for the exponential term (or whether it was even used correctly). It does seem to be the case that, for large gamma, the exponential term is essentially negligible.

Some part of this problem, then, is not solved by finding a mathematical approximation for the exponential term, but in following the physical implications of large drag for the trajectory.

Last edited: Jul 17, 2008