Velocity of a football with and without drag

AI Thread Summary
The discussion revolves around calculating the speed of a football projected straight up, considering scenarios with and without air drag. The first scenario, neglecting drag, leads to a straightforward velocity function based on gravitational acceleration. In the second and third scenarios, where drag is linear and quadratic respectively, participants express difficulty in isolating variables and solving the resulting integrals. The conversation highlights confusion over variable notation and the complexity of integrals involved in the drag scenarios. Ultimately, the participants are seeking clarity on whether the problem's complexity is justified for a homework assignment.
N8G
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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.
 
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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.
 
N8G said:
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).
 
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
 
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.
 
N8G said:
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.
 
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?
 
N8G said:
the integrals for part 2 and 3 both turned out to be horrendous
They shouldn't. What do you get?
 
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