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

[Yosty22]

Homework Statement

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 y_max (given below), show that the speed when the ball returns to the ground is:

v_terv₀ / sqrt(v_ter² + v₀²)

Homework Equations

From a previous problem I worked out, I proved: y_max = (v_ter²/2g)*ln(1+(v₀2[/sub]/v_ter²))

The Attempt at a Solution

I set up the equation of motion as mdv/dt = -cv² + mg. V_ter occurs when the two forces balance, so solving for c in terms of v_ter, I get c = mg/v_ter².

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

dv/dt = g(1-(v²/v_ter²)).

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/v_ter²))

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:

-v_ter/2 * ln(1-(V²/v_ter²)) = 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

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 y_max in for y. However, when they make the substitution, they only have one factor of v_ter in the coefficient before the natural log term even though y_max depends on v_ter². 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 v_ter? If I plug in the y_max equation, I get:

-ln(1-(V²/v_ter²)) = v_ter*ln(1+(v₀²/v_ter²))

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

 

[ehild]

v*dv/dy = g(1-(v²/v_ter²))

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:

-v_ter/2 * ln(1-(V²/v_ter²)) = gy

You miss a square. It is -v_ter²/2 * ln(1-(V²/v_ter²)) = gy

ehild

 

[elegysix]

Solving the integrals, I get:

-v_ter/2 * ln(1-(V²/v_ter²)) = gy

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

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.