Chard wrote:Can we also assume the Earth is flat and gravity is constant and there's no horizontal deceleration at all? If so I get 2h 32m 26.41s. Sadly by that time the actual surface of the Earth is more than 2km below the hypothetical plane the ball bounces along. Of course another answer might be 490 bounces, or 489 bounces and one hitting the ground and rolling.

How did you come up with that time and curvature? Start with a max calculation using the idea that we have the energy for 489 bounces, and that the time it takes to perform half a bounce is 1.4s (from Zach's own calculations), we get t = 489 * 1.4s * 2 + 1.4s = 1370.6s, or just under 23 minutes. But of course only the first bounce will take 1.4s, each will take less time so we expect our answer to be less than that. Where did the 2 hours come from?

For earth curvature, using Yahoo answers (which I know are pretty hit and miss) I found a figure of 8 in/mile, which equates to .1263 m/km in useful units. This means that with your answer of over 9000 seconds, we still only have an earth drop of 23 meters below the theoretical plane. Where did 2 km come from?

Using excel to crunch the numbers, and assuming that the energy loss was only in the Y direction*, I came up with 933.3 seconds. Of course if we look at the curvature of the earth over these 933.3 seconds @ 20 m/s horizontally, it looks like we gain over 2.3 meters of extra height, so curvature of the earth cannot be ignored.

Adding the extra energy gained from the height gain due to curvature of the earth in each step, I came up with 1280 seconds with 648 bounces. Of course this is not exact as the X velocity will slowly decrease as the earth curves and it starts to skew away from being parallel to the ground. I'm not sure if this will help or hurt us (it adds energy in the Y direction, but decreases speed in the X to gain energy due to curvature of the earth) or if it is even a significant different, but I have hit my limit of the math I am willing to do on a webcomic. For now.

*I may end up looking at this again assuming that 1 J of the TOTAL energy is lost. I think there I would assume that the ratio between X/Y energy remains constant over the energy loss (not the ratio of X/Y speed.)