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2. 1D Motion / Kinematics

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Hey, guys, you're going to know how to solve problems where things are moving up or down on the vertical plane, rather on a flat horizontal surface. This is called vertical motion, and in this video we're going to see that vertical motion problems are solved using the exact same steps and the exact same equations as we did for horizontal motion. So let's check it out. But first I actually want to talk about free fall because it's something that your professors and textbooks are gonna cover. We are going to use it later. So the basic idea of free fall is that objects are in free fall if the Onley force that acts on them is the force of gravity. So that's the symbol F G. Even if these things are moving upwards. For example, let's say we had a box and you were holding that box using some string. Basically, you're applying some tension. Well, the other forces acting on it is gravity. It wants to pull it down so you're holding it up using this string, but gravity wants to pull it down, so there's two forces acting on this object. Therefore, it's not in freefall now what if you were to just let the string go while at that point, then the only force that is acting on it is gravity. And obviously the box is gonna fall to the ground. So that means that you have free fall motion. Now, what if instead of dropping the box or dropping string, you were actually just throw the box upwards, basically, just giving it some initial velocity. Well, it turns out that that doesn't really matter, because even though you throw it up once you let it go, the Onley force that acts on it is still gravity. So therefore it's in freefall motion. It doesn't really matter the fact that it's moving upwards, the only thing that does matter is that the Onley thing that's acting on it is gravity. Now, the reason this is important is because objects in freefall experience constant vertical acceleration. So an object in freefall experiences a constant acceleration. And that's super important because that means we can use all of our, um, equations to solve vertical motion problems. So we're gonna use the same list of steps we're gonna use the same exact equations to solve these things. But there are a couple of differences, so let's go ahead and check it out with horizontal motion. We always just drew a diagram like this of our problem. And then we came up with a list of five variables and then from this list, we needed three out of five. So we need three out of five variables in order to be able to pick an equation from this list that didn't have the ignored variable. Well, the vertical motion is gonna be the exact same thing. You're gonna draw the diagram, whether it's moving up or down, whatever. And you're gonna come up with your list of five variables and then basically, from this list here, we need three out of five. So I need three out of five variables, just like I did for horizontal motion to pick a very to pick an equation from this list over here that doesn't have my ignored. So again, the difference really is this use. We use Delta X is for horizontal motion. We're gonna use Delta wise or use Vieques. So now we're gonna use V y. So basically, everything just gets replaced from excess toe wise, and so therefore our equations are instead of instead of having exes in them, they're just gonna have wise Now, the other big difference between vertical motion actually has to do with this acceleration term. This a y. So let's check it out. And basically, what's happening is that objects in freefall in freefall vertical motion are gonna accelerate downwards with a constant vertical acceleration known as the free fall acceleration little G. It's not to be confused with F g. That was the force. This is the acceleration of that force produces. Now that you absolutely need to know about little G is that little G on Earth is always a fixed value of 9.8 m per second squared so G is always equal to negative. Sarriegi is always equal to 9.8, and that's regardless of weight. So that means that if you grab a book and a pencil, and if you drop them, they're both going to accelerate downwards at 9.8 m per second squared. So what does that mean about our acceleration term? Well, our acceleration term is basically just gonna be plus or minus G. So your professors and textbooks are gonna use either a positive or negative sign. That really just depends on what the direction of positive is. And that's based on your problem. So that just means that a wise always gonna be positive or negative, 9.8 m per second squared. We're gonna talk about this in just a second how we actually figure out whether it's positive or negative. So basically, all this means here is that we're gonna use the same exact equations as we did before, but off our variables of our five variables, we already know what one of them is gonna be. It's just gonna be 9.8. We just have to figure out whether it's positive or negative. So let's go ahead and take a look at a problem here. We've got a ball we're gonna drop from rest from a 100 m tall building. So let's just go ahead and draw this out. So I've got a building here. I know this is Ah 100 m, and I'm gonna drop a ball. It's gonna fall, and I know I'm dropping it from rest, which means my initial velocity zero, and I want to calculate the ball's velocity right before it hits the ground. So right at the bottom here, it's gonna have some final velocity v y. And that's actually what I'm trying to find. So this is my target variable. So I've already kind of just drawn the diagram. Let's go ahead and list out the other variables that I need. So I got two of them Be not in view, y So I need delta y that I need a y and then I need Delta t. So these are my five variables here. So what's my delta y My displacement? Well, I'm starting from the top of the building and then I'm heading downwards. So that means that my delta y is gonna be 100 m and then I know my acceleration in the Y direction is gonna be either positive or negative 9.8 m per second. So I've got plus or minus 9.8. Let's let's just say plus or minus G, which is plus or minus 9.8 m per second squared. So remember that this object is falling under free fall acceleration, which always points downwards. It always points downwards. We know the G is always 9.8. The difference is that a Y could be positive or negative, G. So you might have to just throw in a negative sign here. So here's what I wanna warn you guys. Different professors and textbooks are gonna have different ways of figuring out whether this is positive or negative. But in my opinion, it always kind of introduces a limited confusion. So here is a pro tip that's really gonna help you simplify these problems if you always take the up direction to be positive. If we always just say that up is positive, then that means that the acceleration in the Y direction will always be negative. G. It will always be negative. 9.8 m per second squared. So that's what we gonna do here at Clutch A. Y is always gonna be negative. G I Joe is gonna be negative 9.8. And so now that we have our variables here, I've actually got my three out of five. I know my V not zero, and I know my delta Y is now. It's actually not 100 because I've chosen the downward direction to be to be negative. I chose up to be positive and my ball is actually falling downward. So my delta y is actually negative 100 and then my wife is negative. 9.8. So I actually have three out of my five variables. So that means I've got step one and step two. All I have to do is just pick the equation. So let's go to our list. Uh, ignored variable is actually going to be our delta T. Right? That's the one we don't know anything about. So if you go to our list here, uh, this one has t this one has tea, and this one also has t So therefore, our second equation is gonna be the one that we're gonna use. So to figure this out, we're gonna use the second equation. V y squared equals V initial squared, plus to a Y times delta. Why? So this is my target variable here. So let's go to it. So my v y squared is gonna be my v initial is gonna be zero. So right, this is my the initial y pull us to my A y is negative 9.8, and then my displacement delta y is negative 100 so that if you plug this in, you're gonna give you y squared is equal to two. And then if you do this multiplication here, you're gonna get 1960. So now you have to take the square root. So the square root of 1960 and your calculators is gonna give you a number. It's gonna give you 44 0.3. But we have to remember is that when you take the square root of a number, you have to also account for the negative of that number that you get. So, for instance, we're gonna get negative 44.3 m per second. Squared is also a possibility. Because remember, the squared of nine is not just three. It's also negative. Three. Um, so basically, what this means is that we have to figure out which one of these is our correct answer. And so to do that, let's take a look at what we chose to be the direction of positive we chose that was up. So that means that in this positive number here, if we get a velocity that is positive, that means that the that the ball would be moving upwards at this point. If we get a negative velocity, that means that the ball is moving down. So which one of these makes sense? Well, here at the bottom of the path, once you've dropped the ball, obviously it's gonna be falling downwards right before it hits the ground. So that means that the positive number doesn't really make much sense. So therefore, the final velocity of the ball right before it hits the ground is negative. 44.3. Alright, guys, that's it for this one. Let me know if you have any questions.

2

Problem

A rock is thrown vertically upward with a speed of 27.0 m/s from the roof of a 31.0-m-tall building. The rock doesn't hit the building on its way back down and lands in the street below. **(a)** What is the speed of the rock just before it hits the street? **(b)** How much time elapses from when the rock is thrown until it hits the street?

A

(a) vy = 36.6 m/s downward;

(b) t = 3.88 s

(b) t = 3.88 s

B

(a) vy = 11.0 m/s downward;

(b) t = 3.88 s

(b) t = 3.88 s

C

(a) vy = 37.0 m/s downward;

(b) t = 34.2 s

(b) t = 34.2 s

D

(a) vy = 36.6 m/s downward;

(b) t = 6.49 s

(b) t = 6.49 s

3

Problem

A student throws a set of keys vertically upward to her sorority sister who is in a window 14.00 m above. The second student catches the keys 1.50 s later. **(a)** With what initial velocity were the keys thrown? **(b)** What was the velocity of the keys just before they were caught?

A

(a) v_{0y}=37.5 m/s

(b) v_{y}=22.8 m/s

(b) v

B

(a) v_{0y}=16.7 m/s

(b) v_{y}=2 m/s

(b) v

C

(a) v_{0y}=2 m/s

(b) v_{y}=0 m/s

(b) v

Additional resources for Vertical Motion and Free Fall

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