Solving for Position with Zero Initial Velocity

[Bashyboy]

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) v_y(t) = v_{ter}(1-e^{-t/\tau})

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

The Attempt at a Solution

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

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

Through simplification I get:

y(t) = v_{term}t - v_{term} \tau (\frac{t}{\tau}), 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?

 

[haruspex]

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.

 

[Bashyboy]

Oh, thank you.