## [2013-Jan-15] hmm

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Thank you.

### Re: [2012-01-15] hmm ### Re: [2012-01-15] hmm

Soliloquy wrote:
i wrote:It says 'no air resistance', which is different than 'no air'.
If there is air, there is air resistance. Therefore, If there is no air resistance, there is no air.
What he means to say is that 'no air acting on the ball' is different than 'no air on the planet full of humans'

The ball could be shot into an airless tunnel.
God could be pushing the air out of the way so we might better observe his natural laws.
The humans could all be living on a space colony, leaving the earth free for hazardous physics experiments.

### Re: [2012-01-15] hmm

i wrote:It says 'no air resistance', which is different than 'no air'.
If there is air, there is air resistance. Therefore, If there is no air resistance, there is no air.

### Re: [2012-01-15] hmm

I never get tired of seeing pages of pedantry all about the underlying scientific principles of one not very funny joke, no matter how many times I see it.

No, sirree.

### Re: [2012-01-15] hmm

I'd like to think that we all died during a planetwide orgy.

### Re: [2012-01-15] hmm

cosmosis wrote:Just wanted to mention:

Surface area of all land on Earth: 148.3e6 km^2
Surface area of one human (very conservative): 2 m^2

148.3e6 km^2 / 2 m^2 = 7.415e13 people needed to cover the surface of Earth's land.

So the probability of hitting a person, assuming the ball is launched from a random location on land and all dead humans are randomly scattered across the land mass of the earth is: 7e9 people on Earth / 7.415e13 = approximately 0.01% chance of the bowling ball hitting a body on its first bounce. The earth isn't really littered with bodies, and I'd give this answer no credit.
Humans aren't the only things that produce dead bodies when the air goes away.

Granted, though, if everything died where it stood right now, where would any piles be?

### Re: [2012-01-15] hmm

Just wanted to mention:

Surface area of all land on Earth: 148.3e6 km^2
Surface area of one human (very conservative): 2 m^2

148.3e6 km^2 / 2 m^2 = 7.415e13 people needed to cover the surface of Earth's land.

So the probability of hitting a person, assuming the ball is launched from a random location on land and all dead humans are randomly scattered across the land mass of the earth is: 7e9 people on Earth / 7.415e13 = approximately 0.01% chance of the bowling ball hitting a body on its first bounce. The earth isn't really littered with bodies, and I'd give this answer no credit.

### Re: [2012-01-15] hmm

btve wrote:hmm I think the answer is wrong by the way. The time given is the time at which the ball first touches the ground. A study using potential energy should be more usefull here. Sadly I have no time to solve this...
I think you missed the joke here... since the earth is littered with bodies of the dead, they will absorb all the energy of the ball and it will never bounce, hence it is solved with kinematics.
btve wrote:by the way I think that the ball will bounce 1490 times before stoping... (Ek+Ep=1/2mv^2 + mgh).
I agree with you after rereading the question that it is implied that 1 J is lost from ALL the energy, not just energy in the Y direction; although it cleans the number up a bit, I think we cannot make that first assumption. However, an assumption must be made on how much kinetic energy is lost in the X direction vs the Y direction. As I said in my previous post, I think its good to assume the ratio of X/Y energy remains constant from bounce to bounce (as opposed to the ratio of X/Y speed remaining constant.)

Adding those equations to my spreadsheet, we get 1489 bounces over 2838 seconds, or 1838 bounces over 3651 seconds if we take into account the extra energy gained from the horizon "dropping away" as we continue in the X direction. I decided to neglect how the rotating horizon reference will change our X and Y speeds since the angle is so small. Even over the 56 km journey we have only gone 0.14% of the way around the earth, or half a degree.

NOW I'm done (until I decide to actually throw in that rotating reference factor.)

### Re: [2012-01-15] hmm

by the way I think that the ball will bounce 1490 times before stoping... (Ek+Ep=1/2mv^2 + mgh).

### Re: [2012-01-15] hmm

hmm I think the answer is wrong by the way. The time given is the time at which the ball first touches the ground. A study using potential energy should be more usefull here. Sadly I have no time to solve this...

### Re: [2012-01-15] hmm

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.)

### Re: [2012-01-15] hmm

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.

### Re: [2012-01-15] hmm

It says 'no air resistance', which is different than 'no air'.

### [2013-Jan-15] hmm

http://www.smbc-comics.com/index.php?id=2856#comic

I wonder whether I'll hear anything from any future students, who will have read this comic too.... XD