Quadratic and linear drag problem

[Ascendant78]

Homework Statement

Consider the following statement: If at all times during a projectile's flight its speed is much
less than the terminal speed, the effects of air resistance are usually very small. (a) Without reference
to the explicit equations for the magnitude of v t„, explain clearly why this is so. (b) By examining the
explicit formulas (2.26) and (2.53) explain why the statement above is even more useful for the case
of quadratic drag than for the linear case. [Hint: Express the ratio f/mg of the drag to the weight in
terms of the ratio v /vter.]

Homework Equations

(2.26)
vter = mg/b

(2.53)
vter = sqrt(mg/c)

The Attempt at a Solution

I solved part a no problem, but I'm having issues with b. I am assuming when they use "f/mg," since there are no parenthesis around "mg" that g is in the numerator, not the denominator. However, I am not sure what the statement in the hint means. I'm not sure if it is telling me to set things equal to each other (reduce, shift things around, etc.) or if it's telling me to plug values into the other equation. if it were just "vter" in the second value in the hint, then I would assume I plus that in for the force on the first ratio. However, with the information given, I'm at a loss as to what the hint is even suggesting I do.

 

[haruspex]

Homework Statement

Consider the following statement: If at all times during a projectile's flight its speed is much
less than the terminal speed, the effects of air resistance are usually very small. (a) Without reference
to the explicit equations for the magnitude of v t„, explain clearly why this is so. (b) By examining the
explicit formulas (2.26) and (2.53) explain why the statement above is even more useful for the case
of quadratic drag than for the linear case. [Hint: Express the ratio f/mg of the drag to the weight in
terms of the ratio v /vter.]

Homework Equations

(2.26)
vter = mg/b

(2.53)
vter = sqrt(mg/c)

The Attempt at a Solution

I solved part a no problem, but I'm having issues with b. I am assuming when they use "f/mg," since there are no parenthesis around "mg" that g is in the numerator, not the denominator. However, I am not sure what the statement in the hint means. I'm not sure if it is telling me to set things equal to each other (reduce, shift things around, etc.) or if it's telling me to plug values into the other equation. if it were just "vter" in the second value in the hint, then I would assume I plus that in for the force on the first ratio. However, with the information given, I'm at a loss as to what the hint is even suggesting I do.

From dimensional considerations, they must mean f/(mg).
To use the hint, you need the general equation of motion with drag, not just the terminal velocity equation.