Linear Air Resistance with a proof

[aaj92]

Homework Statement

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].

Homework Equations

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

v_{ter}=mg/b = g\tau

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

 

[Curious3141]

Homework Statement

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].

Homework Equations

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

v_{ter}=mg/b = g\tau

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

Do you know the Taylor series for e^x? 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.

 

[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 :)

 

[Curious3141]

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 :)

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.

 

[aaj92]

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

 

[Curious3141]

No worries!