Quadratic drag — a baseball is thrown upwards

[jbeatphys]

Homework Statement

[/B]
A baseball of mass m is thrown straight up with an initial velocity v0. Assuming that the air drag on the ball varies quadratically with speed (f = cv^2), show that the speed varies with height according to the equations.

[Attached]

Where x_{0} is the highest point and k = c/m. Note: x is measured positive upward, and the gravitational force is assumed to be constant.

Homework Equations

The Attempt at a Solution

As I see it, F_netup = F(v_intial) - normal force - drag force(-v) & F_netdown = normal force - drag force.

F_up
(1) Integrate to v(t) (from v0 to 0) and then set v(0) and solve for t, which is the time that it takes to get the peak of the throw.
(2) Integrate to x(t) (from 0 to x0) and then sub in t_{peak}, and solve for v^2.

This is what I have in my head. But I when I complete these calculations I get nothing like what I should be getting [according to the equations] — I get a very complicated equation with hyperbolic trig functions.

Thanks for any help that you can provide.

 

[SteamKing]

Unfortunately, we can't peek inside your head. You'll have to show your work.

 

[jbeatphys]

Unfortunately, we can't peek inside your head. You'll have to show your work.

Okay, I can understand that, not a problem (and probably fortunately).

 

[Chestermiller]

What are the (two) forces acting on the ball when it is rising? What does F_(v initial) mean? A normal force is usually considered a contact force. Is anything contacting the ball as it rises?

Chet

 

[jbeatphys]

Oops. Yes. There is nothing in contact with the ball (other than the fluid, which is air) and so there is no normal force. I meant to say gravitational force, sorry.

The two forces acting upon the ball whilst it rises are the gravitational and the fluid resistance, which are both opposing the initial velocity. You are right to question what is the F_(v initial) it is definitely the wrong conceptualisation of the problem.

To reiterate:

m dv/dt = -mg-cv^2 (up) &
m dv/dt = mg-cv^2 (down) ... I am pretty sure now.

I am going through my calculations again at the moment.

 

[jbeatphys]

Oops. Yes. There is nothing in contact with the ball (other than the fluid, which is air) and so there is no normal force. I meant to say gravitational force, sorry.

The two forces acting upon the ball whilst it rises are the gravitational and the fluid resistance, which are both opposing the initial velocity. You are right to question what is the F_(v initial) it is definitely the wrong conceptualisation of the problem.

To reiterate:

m dv/dt = -mg-cv^2 (up) &
m dv/dt = mg-cv^2 (down) ... I am pretty sure now.

I am going through my calculations again at the moment.

And thanks for correcting my misconceptions.

 

[Chestermiller]

Oops. Yes. There is nothing in contact with the ball (other than the fluid, which is air) and so there is no normal force. I meant to say gravitational force, sorry.

The two forces acting upon the ball whilst it rises are the gravitational and the fluid resistance, which are both opposing the initial velocity. You are right to question what is the F_(v initial) it is definitely the wrong conceptualisation of the problem.

To reiterate:

m dv/dt = -mg-cv^2 (up) &
m dv/dt = mg-cv^2 (down) ... I am pretty sure now.

I am going through my calculations again at the moment.

Much better. Nice job.