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hello everybody today we're going to look at something that's a pretty famous physics thought experiment it's called the monkey gun or the the monkey and the hunter depending on how you've heard of it there are lots of simulations that show this and I just want to say that up front I might have time to show you one myself here but what we're going to do in this video is we're going to actually go through some of the math and show a little bit of a proof we'll call it for how the monkey gun thought experiment works so if you're not familiar with it the way the story goes is that you have a hunter that's sitting over here somewhere and the hunter is trying to hunt this monkey that's up here hanging on a branch the monkey has a trick every time somebody tries to hunt it it watches for a flash of light from the barrel of the gun as soon as it sees a flash of light it lets go of this tree branch that it's hanging off of and it falls to the ground but there's a new hunter in town this hunter knows this monkey's trick so the hunter is trying to anticipate the issue here the hunter knows that as soon as it presses the trigger here that the monkey is going to fall the question is where should the hunter aim the mathematics tells us that the hunter should actually aim directly at the monkey right up here if the hunter aims too high above the monkey obviously that wouldn't work the bullet would go above the monkey often it seems like it would make sense that the hunter should aim lower and try to anticipate the idea that the monkey is going to fall but it turns out that if you aim too low you're going to end up shooting the bullet below the monkey and that's where the mathematics comes in no matter what you should aim at the monkey what's really cool is that the muzzle velocity doesn't matter so let's take a look at the mathematics here well I still have this little pseudo triangle drawn up there's some important information that we might want to draw first and foremost I'm going to use blue here - perhaps signify the monkey we're going to say this is why not meaning the initial height of the monkey there's also an important coordinate over here which is how far away he is I'm going to call this DX that's how far the hunter is from the monkey and then there's some sort of angle here theta that I've already defined that's a displacement triangle so I can deal with vectors that deal with meters or some other length measurement but I would need to come over and also think about the velocity that the bullet leaves the hunters gun so I'd have to draw a separate triangle it would be a similar triangle as far as its proportions are concerned and it's going to have some initial velocity we can talk about the components of the velocity for this particular trajectory for the bullet and we would say that V initial in the X is actually equal to whatever VI is times cosine of theta and then likewise I have a V initial and the Y is going to be equal to whatever the hypotenuse is VI times sine theta and with that I'm ready to set up some functions the general function for dy the height of an object as a function of time is going to be one-half the acceleration times x squared plus V initial in the Y times time plus whatever the initial launch height is the initial and the Y that's a general equation and I'll be able to write one out for both the bullet and the monkey so that's where we have to start I need my functions now I need my y direction for the bullet and the monkey over here I'm going to do a B for bullet in black and I'm going to say D y of the bullet as a function of time is going to be equal to let's say 1/2 acceleration let's go ahead and call that negative 4.9 that's one half of the negative 9.8 meters per second squared that I would have for acceleration negative 4.9 T squared plus the initial velocity times time well the bullet we had that written up here VI sine theta that's the initial y velocity so plus VI sine theta that quantity multiplied by time plus the initial Y location for the bullet I have some ability to define my reference here and I'm going to go ahead and make that bullets initial height the reference location so I'm going to say plus 0 that's nice because I'll go ahead and just get rid of it here so that I can simplify my equation now if I come over and write a similar function for my monkey I find the dy this monkey is just in free fall for the monkey as a function of time is going to be equal to the same negative 4.9 T squared it's the same acceleration due to gravity plus the V initial is zero for the monkey so plus zero I would have a time term in there but since it's V initial and the Y is zero this all goes away plus D initial in the Y remember I define that up here as why not and so I'm going to say Y with a little zero not okay I don't really want that zero in there I kind of want these to be as clean as possible so I'm going to go ahead and rewrite this little spot and just say plus y naught now I'm going to come over and for the bullet let's talk about the X direction in the X direction I'll write it up here D in the X is going to be equal to V in the X x time there's no acceleration up there in the X direction we're not including any wind resistance or anything like that so I don't have an acceleration term basically I'm left with just these two terms except they're going to be in the X direction regardless D X for the bullet as a function of time is going to be equal to the initial velocity in the X Direction is the only velocity in the X Direction is written right here V I cosine theta that full quantity times time plus an initial location except that I'm going to make that a zero also so really both the reference locations in the X and the y are going to be this bullet that means I can come over here and write dx4 the monkey as a function of time is a constant it's just falling straight down so this is going to be equal to this DX quantity that's going out here so I'm just going to call it DX so I got to get some board space here unfortunately I got to get rid of my monkey so now I have my function set up I've got a function for the Y and one for the X both as a function and time for the bullet and for the monkey again this is assuming that you are aiming at the monkey that way the theta that was associated with this problem was the same for both the displacement triangle and the velocity triangle so to really prove that these things are going to Clyde I need certain things to be the same I need time DX and dy to all have the same value at the same time that's the only way I can truly prove that these things would always collide so what I'm going to do first is I'm going to set the dy 's equal to each other as a function of time so that's going to look like this first for the bullet negative 4.9 T squared plus VI sine theta multiplied by time is going to be equal to now the monkeys function negative 4.9 T squared plus y naught this simplifies the process for me you'll notice that that acceleration term was actually the same on both sides and that's actually fundamentally what makes this whole thought experiment interesting so those things go away and now you're left with the simplified version that VI sine theta times time is equal to Y naught this is a good start but now what we have to do is we have to pull more information into this I need to somehow talk about the DX if you're with me on this it is physically possible for both the bullet and the monkey to have the same height but if they're located at different horizontal locations at that time then they certainly haven't collided so somehow we have to pull DX into all of this so over here the monkey is really easy it's always located at DX and so what I want to do is I need to come over here and start talking about the bullet when it's actually at DX so instead of this generic function I'm going to say look when this thing is at DX the same horizontal location as the monkey then I have this expression VI cosine theta times time and if this is going to work then this time that sitting here has to be super related to that time that's sitting up there that's a requirement so let's hang on to this for just a moment and I want to go back up to the y direction I want to manipulate this why not let's get that thing out of there remember I had a triangle that looks like this this was theta this was DX and this was why not well I can write down an expression that says tangent so this is separate here tangent of theta is equal to opposite over adjacent which is y naught over DX so really why not can be written as DX times tan theta so that's how I'm going to write it over here so VI sine theta times time is equal to DX times tan theta okay and I'm going to clear a little bit more board space over here because we're about ready for our very last step if this is all truly equal if all three of these requirements are truly happening if everything is truly equal I should be able to take the time information from the DX and I should be able to plug it in right there so what I'm going to do is I'm going to solve this thing for time and so it's going to look like this time is equal to DX / VI cosine theta and now if I plug that in like I was saying if I plug that in up here then the equation must still hold true it must be equal if I can't make it equal then something is being violated so maybe I'll even write this part in red so you can see all the different parts of it so I'll come over here so I'm just rewriting the black part here VI sine theta times time was DX over VI cosine theta is equal to and then from the monkey side of things we had this expression D X tan theta and if we come look at what we have it's pretty cool the VI cancels with this VI that is the evidence that it doesn't matter what the initial velocity is of the bullet coming out it will still always hit the monkey and then if you recall there's a trig identity that says sine theta divided by cosine theta is actually equal to tangent so you can see that this DX that's right here is actually being multiplied by tan theta because of that trig identity and that is equal to DX times tan theta so that's actually the proof and while it may look like I really needed a lot of board space and a lot of time to do that I was just trying to walk you through all of it it's actually a pretty short little proof that is really just looking at writing the right functions for both the bullet and the monkey and making sure that this is happening so you assume that you have at the same time with the same DX and the same dy in all situations and you start just convincing down your four equations getting rid of other variables until you get this final product that just says hey look this has to be true well that those things are always always going to be equal to each other under these conditions hopefully that made sense to you if it did certainly 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