# Solving for Velocity as a function of height (y) with quadratic drag

1. Sep 13, 2014

### Yosty22

1. The problem statement, all variables and given/known data

Write down the equation of motion for the downward journey of a baseball subject to quadratic drag. Find v as a function of y and, given that the downward journey starts at ymax (given below), show that the speed when the ball returns to the ground is:

vterv0 / sqrt(vter2 + v02)

2. Relevant equations

From a previous problem I worked out, I proved: ymax = (vter2/2g)*ln(1+(v02[/sub]/vter2))

3. The attempt at a solution

I set up the equation of motion as mdv/dt = -cv2 + mg. Vter occurs when the two forces balance, so solving for c in terms of vter, I get c = mg/vter2.

Substituting in for c, cancelling out the mass terms and factoring out a g, I get:

dv/dt = g(1-(v2/vter2)).

Then, dv/dt = v dv/dy (since I want v as a function of y), I can set up the integral.

v*dv/dy = g(1-(v2/vter2))

Then, I can multiply the dy over to the other side and divide everything in the parenthesis over to the left hand side and integrate the left hand side from 0 to V. Since it starts from rest at the top of its journey, V corresponds to the speed of the ball at the bottom when it hits the ground.

Solving the integrals, I get:

-vter/2 * ln(1-(V2/vter2)) = gy

However, trying to solve for V gives me imaginary solutions..

What am I doing wrong here? I checked other online sources for some help to see where I could have possibly gone wrong, and my solution so far agrees with what other online sources have. I went online to look for solutions to where I am getting stuck, and I stumbled upon this site:

http://physics.ucsd.edu/students/courses/fall2009/managed/physics110a/documents/Solution2.pdf [Broken]

If you scroll down to question 2.42, you see them get to the same solution I did (even though I did some more work to get there it seems), but they substitute ymax in for y. However, when they make the substitution, they only have one factor of vter in the coefficient before the natural log term even though ymax depends on vter2. I worked this problem out on my own and got to exactly where they did, but Why could they just get rid of one term of vter? If I plug in the ymax equation, I get:

-ln(1-(V2/vter2)) = vter*ln(1+(v02/vter2))

What happened here or where did I go wrong along the way?

Last edited by a moderator: May 6, 2017
2. Sep 14, 2014

### ehild

You miss a square. It is -vter2/2 * ln(1-(V2/vter2)) = gy

ehild

3. Sep 14, 2014

### elegysix

Should the right hand side be g(y-y0)?
You've got the right ideas of how to solve this.
Usually I have to rework these a few different times until I get it right.
You could also try assuming a solution of some form, my guess would be $y=a-be^{kt}$.
and use y(0)=y0, v(0)=0, a(0)=-mg to find a,b and k.