 This is a proof of the, of a standard physics demonstration problem of called the monkey and the shooter. In this problem, there is a monkey in a tree. A projectile is aimed at the monkey. The question asked to many introductory physics classes is that if the projectile is fired, will it hit the monkey? And most students answer correctly answer no. Because once the projectile is fired, it'll start to experience the acceleration due to gravity. The follow-up question is that if the monkey sees the bullet shot, the projectile shot, and immediately lets go and drops from the tree, simultaneously with when the projectile is fired, will the monkey get hit? And this demonstration is often run in physics classes. And you can see that if the object, the monkey, releases at the same time the projectile is fired, they will hit, they will hit where I just drew that purple star. So what happens is the both objects experience the same acceleration due to gravity. So what our goal here is to show that the monkey and the bullet will end up at the same final position at the same time. So our goal is to show the projectile and the monkey have the same final position at the same time, meaning that they will strike. Okay, so let's set up a few parameters in this problem. Let's say that the monkey is going to start at a position of y initial for the monkey. And let's give that a height of h. And it is going to finish at y final, which is the same for, which will be the same for the monkey or the projectile. Let's say that the projectile is launched with a v initial of v sub i at an angle of theta. Let's give the projectile an initial x and y position of zero. And let's give the projectile a final horizontal position. So that's final horizontal position for the projectile. And let's just simplify that this distance traversed, let's call that x. So there's our scenario for our monkey. As in solving all projectile problems, it's really good to organize what we know into some organized format in order to ensure that we can keep all of these different variables in order. Also, since projectiles are vectors, we need to make sure that we choose a positive coordinate system for these factors. I'm going to choose a coordinate system such that up is positive and to the right is positive. So I'm going to open up a fresh page here and so that we can write down what we know for each of these. So for the, we will start with the monkey and write down our knowns for the monkey. We know that the monkey is starting at a position of h and is finishing at a position of y final. We know that the monkey starts from rest and is going to accelerate downward for the amount of time until it reaches this final position due to the force of gravity with the acceleration due to gravity. Okay, so we know that this y final position and this time will be the same for both the monkey and the projectile. Now let's write down what we know about the projectile. I'm going to separate it out into its components of vertical and separate it into its vertical and horizontal components to try to keep it organized. In the horizontal direction, we know that the monkey starts at a position of zero and it will end up with some x final position, xfp, and that that distance traversed will be x. We know that it will have an initial velocity that is a component of this initial velocity at this angle and we know that it will be in the air for the same amount for the time t. So let's write all those down. Okay, I also forgot to mention that horizontally our projectile does not accelerate. There are no forces changing its motion horizontally is only the acceleration due to gravity that keeps this thing that changes the motion of this projectile. Okay, so I can write it in my table. I want to focus in a little bit more on this initial launching velocity. So if you remember from when we talked about our projectiles, let me, all right, if we have an object launched at an angle, it is going to have some portion of its motion that is going to be horizontal and some portion of it that is going to be vertical. If our launching angle is theta, then this will, our red portion will be our v initial in the vertical direction, our blue portion will be our v initial in the horizontal direction and the firing speed will be the hypotenuse of this vector triangle. If we want to write down, so as you can see, the v initial in the vertical is related to the launching speed v initial and the angle theta by the sine function and the horizontal piece is related to the launching speed and the angle by the cosine function. Now we can include these in our information of known values. In our vertical direction, we know that this projectile, that the projectile is going to accelerate rise and fall due to the acceleration due to gravity for the same amount of time as it is traveling horizontally. It will start its position at zero and end up at this x final and it's going to be have a vertical initial velocity as dictated by our vector triangle. So let's go ahead and write all that information down. So now that we have all of this information written down for our values, it's time to start seeing how we can pull it together. If you recall, our goal is to show that the monkey and the projectile have the same final position at the same time. So let's start by looking at the monkey and finding an expression for its final position in terms of the things that we wrote down that we know. So here's our information for our monkey. I'm just going to go ahead and move some of this around so we have some space for working. Okay. So for our monkey, we are going to look at these variables and see which equation of kinematics will relate them all. And we have acceleration v initial, delta, displacement, and time. So the equation that links all those together is that displacement is equal to v initial t plus 1 half a t squared. All right. Let's substitute what we know. Our delta y is our y final minus h. The v initial is 0. Our acceleration is minus g. Let's substitute those in. Since v initial was 0, that term does not have to be included. And the negative in front of the 1 half term there is due to the fact that we chose down to be the negative direction. So our acceleration is negative. I'm just going to get the h to the other side and we will have an expression for our y final. So now let's put this aside. If we can show that the final position, vertical position in the same amount of time for the projectile is the same, then we can, then we have shown that these objects will meet at the same location at the same time. So let's go ahead and figure this out with our projectile. What we're going to do is look at both the horizontal and vertical components of this motion independently, then combine them together. So let's go ahead and write an expression for both our, for our horizontal motion of the projectile. We know a, t is common. We have an expression for displacement and for v initial. Let's go ahead and pick out our equation that combines those. It's going to be the same one involving displacement time, acceleration and initial velocity. Our horizontal acceleration is zero. So this will simplify down a fair bit. Let's do that, let's plug in what we know and simplify. Now we're going to do the same thing with our vertical motion. Right? We have the same variables, so we are going to use the same equation of kinematics that ties all them together. And let's substitute, write that down. We'll substitute in what we know and then simplify. So you might be looking at this and thinking, well, that really doesn't look like what we found for our final position for the monkey. And it doesn't yet. But let's make a few substitutions for things that we know and see where we can get. As you can see, what I'm going to do is I'm going to solve this purple equation here for time and substitute that for the t here. Now I'm going to substitute that in, that in for this t in our final position expression. Now I'm going to simplify this expression. We can see that the v initial terms will cancel out and sine over cosine will simplify to tangent of theta. So looking at this, you might be thinking, that's still not quite what I want it to look like. But this is an expression of position. And this is an expression of position. Let's see if we can relate them using trigonometry. Let's go back to our picture that we started with. We said that this right triangle where x is this horizontal displacement of the projectile, h is that starting height of the monkey, and those form two sides of a right triangle with the angle theta. Let's draw that. As you can see from this triangle, now that we drew it with a little bit more clarity, that these, this height h and this distance x are related by the angle theta. They are opposite and adjacent to the angle theta. Therefore, they are related by the tangent function. Or if we rearrange this equation, we see that h is equal to the product of the tangent of our theta. Times our distance x. All right. Well, let's see how this helps us. If we look over here, we see that tangent of theta times x appears here. So we know that another way to write that is as h minus 1 half g t squared, and as we can see, that final position is the same as we got for the monkey. Same final position in the same amount of time. So if our monkey lets go at the same time that the projectile is launched, then they will hit at the same location at the same time. So we have shown a solution to this problem. This is not the only way to solve it, but it is one way. Thank you.