Velocity of a football with and without drag

[N8G]

Homework Statement

A professional thrower projects a football straight up in the air.
1. Assuming there is no air drag on the football, find the speed of the football as a function of height as the ball goes up.
2. Assuming the air drag on the football varies linearly with speed, find the speed of the football as a function of height as the ball goes up.
3. Assuming the air drag on the football varies quadratically with speed, find the speed of the football as a function of height as the ball goes up.

Homework Equations

F=ma which extends to mv dv/dx through chain rule
linear drag = c1v
quadratic drag = c2v^2

In each case the sum of the forces in the y direction equals mg minus the corresponding drag term

The Attempt at a Solution

1. mg = mv dv/dx ... separate variables, v(y) = root( 2gy +vo^2)
2. mg - c1v = mv dv/dx ... unsure how to isolate variables, haven't made it to part 3 but same issue

I'm hoping that I'm missing something simple that I've just overlooked, any help would be greatly appreciated.

 

[mjc123]

Are you confusing x and y, or using them for the same thing?
If up is the positive x (or y) direction, the gravitational force is -mg, not mg.

 

[haruspex]

mg - c1v = mv dv/dx ... unsure how to isolate variables

Try a bit harder... it really is very simple. You just want all the references to x on one side and all the references to v on the other.
And as mjc123 mentions, you should check your signs. Or maybe you are taking g to have a negative value (which is a valid approach).

 

[N8G]

My bad, any x's should be y's.

For the second part of the problem I have:
mvdv/dy = mg - cv which I reduce to
dv/dy = g/v - c/m
From here I don't see a way to isolate the v term on the rhs from the dy when separating my variables.

And I am taking g to be -9.8m/s^2

 

[N8G]

Never mind. I think I lost my mind and forgot about how division works.

I should be able to just say:

mg - cv = mv dv/dy

1 = mv/(mg-cv) dv/dy

dy = mv/(mg-cv) dv

Sorry for that.

 

[haruspex]

Never mind. I think I lost my mind and forgot about how division works.

I should be able to just say:

mg - cv = mv dv/dy

1 = mv/(mg-cv) dv/dy

dy = mv/(mg-cv) dv

Sorry for that.

Glad to see you found your mind.

 

[N8G]

That being said, I figured out how to separate the variables but the integrals for part 2 and 3 both turned out to be horrendous given that I’m looking for the velocity wrt height functions. Each integral needed either aggressive attempts at u substitution or partial fraction decomposition followed by an annoying transform. I find it hard to believe thay professor intended that much work for a minute 10 pt homework assignment. Am I missing something elementary that would make my life easier?

 

[haruspex]

the integrals for part 2 and 3 both turned out to be horrendous

They shouldn't. What do you get?