# Line Integral of Force of Air Resistance

So this is a Physics problem, but what I'm mostly hung up on are the integration techniques used.

The question is this:

A baseball (radius = .0366 m, mass = .145 kg) is dropped from rest at the top of the Empire State Building (height = 1250).

I solved parts a) and b) quite easily. Enter part c):

Calculate the total energy dissipated by the falling baseball's initial potential energy and its final kinetic energy. (Hint: In part (c) make approximations when evaluating the hyperbolic functions obtained in carrying out the line integral.)

I've attached a picture of the solution manual's solution, which I understand only to the point to see that they're using integration techniques I have no knowledge of.

So what is going here? I've never seen hyperbolic functions used up until this point. I can see that $v^3$ results from converting $v^2$ from a position function to a time function. Why $v^3$ turns into $(-v_t tanh{1 \over \tau})^3$ is anyone's guess in my book, and I continue to get exponentially more lost from there.

I'd appreciate any help anyone can offer. I'm not finding all that much on this on the web (maybe I suck at searching) and my professor generally seems pretty unwilling to explain things.

Here's my take on this... The total energy involved in the dissipation of the ball is

$E = \int F dx$ as you have there, which we then get:

$= -\int c v^2 dx$.

Now, we know that $v = dx / dt \Rightarrow dx = v dt$, so, we get that the integral becomes:

$-c \int v^3 dt$, by a simple substitution.

Now, where does the $tanh$ function come from? You are doing a air drag question. The forces involved let's say are Newtonian gravity, and the air resistance (which is proportional to the speed of the object squared). Applying Newton's second law, we get that

$m a = mg - c v^2$,

since $a = dv / dt$, we have that

$mg - cv^2 = m \frac{dv}{dt}$.

This is an ordinary differential equation which one can solve for $v(t)$. I don't know how much ODEs you have done in your studies, so I will simply tell you the solution:

$v(t) = \frac{\sqrt{g} \sqrt{m} \tanh \left(\frac{\sqrt{m} \tanh ^{-1}\left(\frac{\sqrt{c} \text{vo}}{\sqrt{g} \sqrt{m}}\right)+\sqrt{c} \sqrt{g} t}{\sqrt{m}}\right)}{\sqrt{c}}$.

This assumes something specific about the initial conditions, and is a bit different than what you have, but the basic idea is that $tanh$ comes into play from solving the air drag ODE from Newton's 2nd law.

The rest just follows from integrating the hyperbolic tangent function, here is a helpful link for that: http://math2.org/math/integrals/more/tanh.htm