Calculating Terminal Velocity with Air Resistance: A Case Study in Free Fall

[eehiram]

My textbook source is:
Fundamentals of Physics, 6th edition, by Halliday, Resnick, Walker

According to Newton's well known 2nd Law of Motion:
F_net = ma

In chapter 2, in the case of free fall, the F_grav = mg,
where g = -9.8 m/s², assuming that movement along the axis of y is positive going upward from the Earth's surface.

However, in order to make a slighter more elaborate calculation, let us attempt include to incorporate air resistance, as in chapter 6:

If we have the following values to insert into the Drag equation:

Mass density of air = ρ = 1.29 kg / m³

Object speed in m/s = v

Drag coefficient = C_drag = needs to be looked up or calculated

Effective Cross-sectional Area in m² = A



The drag equation in chapter 6, section 3 appears to be:

F_drag = (1/2) ρv²C_dragA


As air resistance increases with v², the F_drag reaches a value equal in magnitude and opposite in direction to F_grav.

Then terminal velocity might be attained, and the object's free fall may cease to accelerate.

Terminal velocity v_terminal can be solved for by calculating the case of F_net = 0 = F_drag + F_grav

F_drag = -F_grav
(1/2) ρv²C_dragA = -mg
etcetera...

Is this correct? This is not a homework assignment question.

 

[jtbell]

Notice that if you try to solve your last equation for v, you'll end up taking the square root of a negative number. You can get rid of that pesky minus sign by going back to your net force equation and recognizing that it's a vector equation:

$$\vec F_{net} = 0 \\
\vec F_{drag} + \vec F_{grav} = 0 \\
\vec F_{drag} = - \vec F_{grav}$$

The last equation says that the two forces are opposite in direction (minus sign) and equal in magnitude:

$$F_{drag} = F_{grav}$$

This equation is for the magnitudes, so we drop the minus sign.

 

[eehiram]

Yes, you are right. I should have used vectors in my initial post. Thank you for correcting my mistake.