Solving for Velocity of a Falling Skydiver: m(dv/dt)=mg-kv2

[Jamin2112]

Homework Statement

The velocity v(t) of a skydiver falling to the ground is governed by

m(dv/dt)=mg-kv²,

where m is the skydiver's mass, g is the acceleration due to gravity, k > 0 is the drag coefficient, and v(t)≥0.

(a) Solve this equation for v(t) with the initial condition v(0).

Homework Equations

The Attempt at a Solution

Divide m throughout

==> dv/dt = g - (k/m)v²

Factor out a k/m

dv/dt = k/m (mg/k - v²)

Separate

dv / (mg/k - v²) = (k/m)dt

Now I have it in the form ∫ dx/(a²-x²) = (1/2a)ln[(a+x)/(a-x)], with a = √(mg/k) and x=v, obviously.

==> [1/√(mg/k)] ln[ (√(mg/k) + v) / (√(mg/k) - v)] = (k/m)t + C

..... and this all seems far too complicated.

Suggestions, please.

 

[Hellabyte]

I think you have to realize that as you get into upper division stuff such as this the answers to problems are going to get messier. you are right that the +v -v thing is intimidating, but that doesn't mean its wrong. Also, why didn't you try and solve it for v? The right answer could have popped out. I remember doing this differently so ill just go through that method. I looked up the integral and found this:$$\int \frac{dx}{a^2 - x^2} = \frac{1}{a}ArcTanh(\frac{x}{a}) + C$$

This is the same thing that you got but of a different form because

$$ArcTanh(x) = \frac{1}{2}ln(\frac{1+x}{1-x})$$

But now we have a scary looking ArcTanh part. When you see the Arc part remember that we could just take Tanh of it to get whatever is inside the function:$$Tanh(ArcTanh(x)) = x$$

in physics whenever you see scary trigonometric(or hyperbolic in this case) that you want to get rid of, look for identities to get it in terms of better functions.

look at this page for some help for the rest of the way.
http://en.wikipedia.org/wiki/Hyperbolic_function