**Annotation category:**

Chapter 2

Note: |

As is often the case with back-of-the-envelope problems, there are a number of ways to approach this. Here's the one I used:

The pitcher's mound is 60 feet, 6 inches from home plate = 18.44 m. I guessed a baseball was about 3 inches in diameter and weighed about 6 ounces; I subsequently looked it up on the web and found the NCAA requirement is "The ball shall have a weight of 5.12 +/- 0.035 oz. The circumference of the baseball shall be 9.05 +/-0.05 in." (I was delighted to see that the NCAA recognizes the concept of measurement (and manufacturing) uncertainties!) So my guesses on my napkin were within ~15% in both cases, but I will use the real numbers since I now have them (5.12 oz = 145.2 gm and 9.05"/π implies r = 3.66 cm).

Now, it was raining fairly lightly, and I estimated it would take at least 4 or 5 hours to accumulate an inch of rain at that rate; I choose 0.5 cm/hr to use in the calculation. This means that in an hour, a layer of water 0.5 cm thick would cover the ground (assuming none sunk in or ran off -- that's how much is actually falling through the air, some of which hits the baseball). The ball covers its trip quickly: 95 miles/hr x 1609 m/mile x 1 hr/3600 s = 42.46 m/s, meaning it covers the 18.44 m from the pitcher's mound in 0.43 sec. At any given moment, it is obscuring (and therefore hitting or being hit by)
raindrops over an area of πr^{2} = 42.0 cm^{2}. Now in an hour, this much
ground would accumulate 42 cm^{2} x 0.5 cm = 21cm^{3} of water. In 0.43 s, then,
it will intercept 21 cm^{3}/hr x 1 hr/3600 s x 0.43 s = 0.0025 cm^{3} and, since
water has a density of 1.0 gm/cm^{3}, the ball encounters 0.0025 gm of water during the pitch. This is only 0.0025gm/145.2gm = 1.7 x 10^{-5} gm of its original mass, or a 0.002% effect. We could get fancier by talking about where the water hits and the resultant momentum transfer, etc., but it is clear from getting this far that the effect is completely negligible.