# Linear Air Resistance need help with a proof

1. Jan 24, 2012

### aaj92

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

Equation 2.33 gives the velocity of an object dropped from rest. At first, why v$_{y}$ is small, air resistance should be unimportant and 2.33 should agree with the elementary result V$_{y}$ = gt for free fall in a vacuum. Prove that this is the case. [HINT: remember the Taylor series].

2. Relevant equations

equation 2.33---> v$_{y}$(t) = v$_{ter}$(1-e$^{-t/\tau}$)

v$_{ter}$=mg/b = g$\tau$

3. The attempt at a solution

I substituted g\$\tau$ for v$_{ter}$ uhhhh and yeah... not sure what to do? I know it's probably really simple but I'm stuck

2. Jan 24, 2012

### Curious3141

Do you know the Taylor series for ex? Just put x = -t/τ and expand the exponential term to the first few terms. Terms with the power of t greater than one (i.e. t2 and higher) can be disregarded because we're talking about a small time t (after release) here.

Once you do the simple algebra, you'll get the expression you need.

3. Jan 24, 2012

### aaj92

yeah the taylor series is e^x = 1 + x

so I just make it e^(-t/tau) = 1 - (t/tau) ? then do i substitute for tau? sorry I'm just not seeing v = gt :/

wait so then you plug in 1- (t/tau) into the original equation and then solve it right? haha sorry. I think i figured it out. thank you :)

4. Jan 24, 2012

### Curious3141

Just a couple of things: $e^x = 1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + ...$ (an infinite series).

$e^x \approx 1 + x$ is an approximation that's only valid for small x. So you can use it here. However, it's wrong to simply state $e^x = 1 + x$ like you did. Remember that this is only an approximation where you're ignoring all those higher powers of x (because they're too tiny).

Also, you're not solving an equation, just simplifying an expression. But yes, put (1-t/τ) into that expression in place of $e^{-\frac{t}{\tau}}$, put $v_{ter}$ as gτ, and you'll quickly get the result.

5. Jan 24, 2012

### aaj92

haha yeah it was way easier than I was trying to make it :p thank you so much!! :)

6. Jan 24, 2012

No worries!