Back-of-the-envelope calculations are often very useful but not when they are an order of magnitude wrong. Mahajan^{1} used such calculations to predict that a 100 lb iron ball dropped from a height of 50 m would hit the ground about 10 m ahead of a 1 lb ball. Mahajan improved the calculation with a “few manipulations” and came up with a more plausible answer of about 80 cm, indicating that Galileo's claim of a 5 cm difference suggests that he did not actually perform such an experiment. I have three comments. First, I doubt that anyone would have sufficient strength to hold a 100 lb ball in one hand at arm's length in order to drop it at the same time as a 1 lb ball in the other hand. Second, an 80 cm difference corresponds to a time difference of only 0.025 s when the balls hit the ground at about 31 m/s. I doubt that anyone could drop two balls simultaneously within that time frame, and at zero speed, using hand-eye coordination alone. However, my main comment concerns the 10 m back-of-the-envelope calculation. It is obviously incorrect, but why?

Mahajan argues that the mass of the air swept out by a ball divided by the mass of the ball is probably equal to the fractional change in the height that the ball falls due to air resistance. Maybe it is, maybe it isn't. Mahajan does not elaborate. However, it is easy to see what is wrong with that assumption. That is, air does not behave like a solid. It behaves more like a liquid and flows around the ball from the front to the back as the ball falls through the air.

In a vacuum, both balls would hit the ground with *v* = 31.305 m/s after 3.194 s if released from rest. If the drag coefficient for both balls is 0.5, then the 1 lb ball hits the ground with *v* = 30.4 m/s after 3.225 s and the 100 lb ball hits the ground at *v* = 31.12 m/s after 3.200 s.^{2} The loss in momentum by the 1 lb ball is 0.410 kg m/s. The loss by the 100 lb ball is 8.392 kg m/s. If the momentum was transferred to all the air in front of the ball, then the increase in the air speed would be only 3.74 m/s for the 1 lb ball and 3.56 m/s for the 100 lb ball. Consequently, the ball does not collide with the air in the same way that it might collide with a rubber ball having the same mass as the air. Alternatively, the fractional loss in kinetic energy of the 1 lb ball (compared to the same ball dropped 50 m in vacuum) is 0.057, while the mass of the air ahead of the ball is 0.241 times the mass of the ball itself. Regardless of how the problem is treated, the first step is to get the physics approximately right. In the present case, a reasonable approximation might be to assume that each ball collides with a small fraction of the air ahead of it, given that most of the air sneaks around the back of the ball rather than being projected forward.