Linear Air Resistance

In summary, the conversation discusses equations that give the velocity and position of an object dropped from rest, and how they relate to free fall in a vacuum. The solutions to two parts of the problem are given, and the person asking the question is seeking help with finding a solution to the second part. It is determined that this involves taking further terms in the Taylor expansion to avoid getting a result of zero.
  • #1

Homework Statement

Equation (2.33) gives the velocity of an object dropped from rest. At first, when Vy is small, air resistance should be unimportant and (2.33) should agree with the elementary result Vy = gt for free fall in a vacuum. Prove that this is the case. (b) The position of the dropped object is given by (2.35) with V0y = 0. Show similarly that this reduces to the familiar y = 1/2 gt^2

Homework Equations

equation (2.33) [itex]v_y(t) = v_{ter}(1-e^{-t/\tau})[/itex]

equation (2.35) [itex]y(t) = v_{ter}t + (v_{0y} - v_{ter}) \tau (1 - e^{-t/\tau})[/itex]

The Attempt at a Solution

I was able to solve part a. Here is my attempt at finding a solution to part b:

[itex]y(t) = v_{term}t - \v_{term} \tau [1 - (1 - \frac{t}{\tau})][/itex] I used the first two terms of the taylor seiries to approximate the exponential function.

Through simplification I get:

[itex]y(t) = v_{term}t - v_{term} \tau (\frac{t}{\tau})[/itex], which is clearly going to be zero--but that does not make sense, for it implies that the position is always zero for all time.

What did I do wrong?
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  • #2
You are neglecting terms in t-squared and higher. Since the answer is like t squared it is inevitable you will get 0. You need to take further terms in the Taylor expansion.
  • #3
Oh, thank you.