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Standoff

Two Cannons

Let's say we have two cannons aimed directly at one another ( as my horrible attempt shown in the image ).

enter image description here

The Angles The cannon on top is aiming down towards the one on bottom right, and of course, the bottom one pointed upwards. They are pointed towards each other as exactly as can be.

Resistance It happens to be the calmest day in recorded history. There is no air resistance at all.

Coordinated Shots Let's assume both cannons are fired at the exact same time, exact same speed and power. Yes, aimed at each other.

What will happen?

Note: Before considering tagging this puzzle "too vague", or "not enough information provided", which is at first-glance a reasonable assumption - I just want to assure you that there is a consistent outcome.

Hint: Read up on this topic if you get stuck:

fixed-point theorem in topology

Good Luck. I'm looking forward to seeing some work written out. I have the answer fully documented when it's time to post it.

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    $\begingroup$ If you point a cannon down, won't the cannonball just roll out? $\endgroup$ – Bass Apr 8 at 6:16
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    $\begingroup$ Air resistance does not disappear when the wind dies. $\endgroup$ – A. I. Breveleri Apr 8 at 14:23
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    $\begingroup$ @A.I.Breveleri Yeah, when I think "calmest day in recorded history" I don't associate that with "all the air completely disappeared". I get what they were going for... but still. $\endgroup$ – JMac Apr 8 at 19:09
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    $\begingroup$ Even seeing the answers, I still find this extremely vague. It seems weird that you're just expected to conclude that the cannonballs hitting is special and is what you were looking for. At a bare minimum you should ask "what will happen to the cannonballs?" (though that still seems sufficiently vague to me). $\endgroup$ – JMac Apr 8 at 19:13
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    $\begingroup$ @Bass depends on the cannon type. Modern cannons shells are sized to have some resistance as they move down the barrel, to minimize gasses blowing by the shell and reducing muzzle velocity. Old frontloaders made before accurate machine tools had larger tolerances, so you'd have a shell with some rag around it to create the friction. $\endgroup$ – Hobbes Apr 9 at 17:37
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Alain Remillard has given the mathematician's answer. Here's the physicist's one:

Step 1:

assume gravity does not exist.

Obviously, in such a universe, regardless of their speed, the cannonballs will travel in a straight line and hit each other in the middle.

Step 2:

Assume "Step 1" does not exist.

A uniform gravity field will pull all masses at the same acceleration, regardless of the mass or its speed. So, if we let our coordinate system fall with the gravity, we have a situation that looks exactly like Step 1. (Except that the cannons and the ground seem to be "falling upwards".)

Therefore

The cannonballs will collide unless they hit the ground before reaching each other.

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    $\begingroup$ ^vote with a note: For the gist of the result you make it sound like the balls don't even need to be shot at the same velocity as long as they point at each other.(And the image you evoked of the cannons was good for a chuckle.) $\endgroup$ – humn Apr 8 at 7:00
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    $\begingroup$ @humn, yes, that seems to logically follow! Also, please note that I do not recommend (or for that matter, condone) jumping off the castle wall in order to create a stationary observer for Step 2. (Unless the gravity is pointing into some more interesting direction than the hard, unforgiving ground, which also shouldn't affect the result) $\endgroup$ – Bass Apr 8 at 7:06
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    $\begingroup$ Maybe I should add a word of caution about playing fast and loose with coordinate systems like this: while you can, in general, have your coordinate system moving at a constant linear velocity without affecting the laws of physics in any way, here we have an accelerating coordinate system. That means that anything that's not in freefall will be gaining a "magical acceleration out of nowhere" in this system, which makes it crucially important that the cannons are fired at the exact same time. $\endgroup$ – Bass Apr 8 at 7:33
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    $\begingroup$ Gravity will not affect the outcome . The event will still happen, however, the coordinates of the event will change whether gravity is present or not. Just an interesting FYI. $\endgroup$ – John S. Apr 8 at 14:02
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    $\begingroup$ Instead of two cannonballs, I've seen this experiment / demo done in real life with a ball bearing "bullet" and a target that drops at the same time the shot is fired. (e.g. in a first year physics class.) This demo is traditionally called "monkey and hunter" (wikipedia). You can google monkey hunter gravity ball bearing to find video demos. So yes, @humn, no need for the initial velocities to be opposite, just on a collision course if there was no gravity. $\endgroup$ – Peter Cordes Apr 8 at 18:11
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They will

Collide in mid air, exactly at the middle.

I did it mathematically

Suppose the horizontal distance between both cannons is $d$ and the up angle from right cannon is $\theta$. Then, the left cannon is at a height of $d\tan\theta$ and aim down at an angle of $\theta$. Since the horizontal speed of the cannonballs are the same, there is à time when they are both at the same horizontal position. It is left to prove they are at the same height. Position for both cannonballs are given by $$\begin{cases}x_l=tv\cos\theta\\y_l=d\tan\theta-tv\sin\theta-\frac12gt^2\end{cases}\qquad\begin{cases}x_r=d-tv\cos\theta\\y_r=tv\sin\theta-\frac12gt^2\end{cases}$$ Horizontal positions are the same at $$x_l=x_r\implies > t=\frac{d}{2\cos\theta}$$ Vertical positions are the same at $$y_l=y_r\implies > d\tan\theta-tv\sin\theta-\frac12gt^2=tv\sin\theta-\frac12gt^2\\\implies > d\tan\theta-tv\sin\theta=tv\sin\theta\\\implies > t=\frac{d\tan\theta}{2\sin\theta}=\frac d{2\cos\theta}$$

Then they are

at the exact same position and they collide at the middle.

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    $\begingroup$ If anyone could hide my explaination in a spoiler tag, it will be appreciate. M'y phone won't let me do it. $\endgroup$ – Alain Remillard Apr 8 at 6:24
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    $\begingroup$ Your explanation is spot on. You explained this puzzle clearly, and I am hoping your visuals will help new physics students understand this principle. Thank you for contributing such useful content. $\endgroup$ – John S. Apr 8 at 14:00

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