|MadSci Network: Physics|
There are several previous answers in our archives dealing with terminal velocity, which I suggest you read. Use our search engine to search on "terminal velocity" with "find the exact phrase" turned on.
At least one of the previous answers refers to a page on the Web that allows one to calculate the terminal velocity of spheres, knowing the size and density. The terminal-velocity calculator is found here.
I found the pertinent information about the specifications of golf balls at this site. There we find that a golf ball is supposed to be no more than 45.93 grams in mass and no smaller than 42.67 mm in diameter. Using these as nominal dimensions in the calculator above we obtain a terminal velocity of 31.24 meters per second. [There is one fine point that one needs to know in order to be able to use the calculator. It does not directly accept a value for the mass; the mass is a computed value from the radius and density. So, we need to calculate the density. The volume of a golf ball of radius 2.1335 cm is 40.6786 cm3, and since the mass is 45.93 g the density must be 45.93 g / 40.6786 cm3 = 1.12909 g/cm3. Put that value into the density textbox.]
Now, what you really want to know is how far the golf ball has to fall before it reaches that terminal velocity. In a previous answer that I wrote for MadSci I worked on an Excel spreadsheet that calculated, using numerical simulation, the trajectory of shot from a shot shell. Your problem is actually easier because we only have to consider one dimension. So, I altered the spreadsheet to calculate only the y velocity and distance, using the appropriate size and mass for a golf ball. If you look at the data and the graphs you will see that the velocity is getting very very close to the terminal velocity after about 10 seconds, and by that time the golf ball has fallen only about 244 meters. In your question you were wondering about the difference between 1 and 2 miles high. It turns out that the terminal velocity is reached in only about 0.15 miles!!
However, you may have noticed that the velocity after 10 seconds of falling is only 31.136 meters per second, while the terminal velocity found in the terminal-velocity calculator is 31.24 meters per second. How long does it take for the golf ball to actually get to 31.24 meters per second? Although the correspondence between the terminal-velocity calculator and the spreadsheet is not perfect (Notice that the velocity after 20 seconds of falling, in the spreadsheet, is 31.25 meters per second!), in the spreadsheet we find 31.24 meters per second at 13.63 seconds. What is more important, though, is that after 20 seconds of falling the velocity is changing in only the 8th significant digit!!
Okay, so let's say the terminal velocity is reached in 20 seconds instead of 10 seconds. At that time the golf ball has fallen 556 meters, which is still only 0.346 miles, which is still much less than the difference you suggested.
You have to decide how close to 31.24 (or 31.25, in the spreadsheet) meters per second the golf ball is falling in order to "qualify" for being "at terminal velocity". According to the spreadsheet, at 10 seconds of falling the velocity is within 0.36% of the "real" terminal velocity. Is that close enough? How about at 20 seconds, where the change from one interval to the next is in the eighth significant digit? Is that close enough? If not, the interval in the spreadsheet needs to be smaller, and the maximum time over which the digital simulation is run needs to be longer. The amount of memory available to Excel is the most important limiting factor. Have fun!!!
John Link, MadSci Physicist
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