Hey, guys. So now we've been introduced to the various kinds of projectile motion, it's time to start solving some problems. In this video, I'm going to give you a great system for solving any one of your projectile motion problems that you might see. We've got a list of steps and equations. Let's just get right to it.
We've got a ball that rolls horizontally off of a table and we know the height and the speed. We're going to calculate this first part, the time it takes for the ball to hit the ground. Here's how these problems are always going to go. Your first step is to draw the paths in the x and y axes. We've got this ball that rolls horizontally off the table. First, we need to sketch out what the trajectory is. Once it leaves the table, it's going to be under the influence of only gravity. It's going to take a curved parabolic path. We're going to draw the path in the x and y axes.
What happens is even though this is a two-dimensional path from here to here, we can break this up and imagine if we could only move along the x-axis that we'd be moving from here to here, and in the y-axis, we've just been moving from here down to here. So now the points of interest. Points of interest could be anything like your initial, so where your starting position is, that's always going to be point A. And then your final, so the final is going to be down here and then anything else that may happen in between like you reaching a maximum height or something like that. In this particular example, we're only told that we're going from the table down to the ground. So that's my initial and final. So I'm just going to use those two points of interest, A and B.
The second step is we need to determine the target variable. In this case, we're looking for the time it takes for the ball to hit the ground. So that's going to be t. The third step is which interval are we going to use? An interval is basically just between what two points in the diagram or in the problem are you looking at. For example, when we go pick our equations here, if we're using an equation like Δx=vxt, this is a displacement. But between what two points, what interval are we looking at for that displacement?
In this particular problem, because we're looking for the time it takes for the ball to go from the table at point A to the ground at point B, the interval we're going to use is the one from A to B. Now what equation do we use to solve for tAB now that we know the interval? Well, if you take a look at our equations, we have time in the x-axis, but time also appears in the y-axis. In fact, in all your projectile motion problems, time, this variable, can be found by either the x or y-axis equations. Here's the deal. We have four equations in the y-axis, but we only have one equation to use in the x-axis. And because the x-axis only has one equation, we're always going to try the x-axis first because it's the simplest thing to do. In the x-axis, we've got Δx=vxt from A to B. So this is our target variable. We just need to know the other two variables here.
What about the displacement? The displacement is going to be from point A to point B. What about the x velocity? Well, we have an initial velocity. We're told that the initial velocity of the ball is just 3 meters per second. So in general, if we want the x component and the y component, we just have to use our vector's equations, our cosine and sine functions. So my initial velocity in the x-axis, which I'm going to call vax because it's the x velocity at point A, is just going to be v0·cos(θ), and it's going to be 3·cos(0) which is 3. In the y-axis, we're going to do the same thing vay=v0·sin(θ), so it's going to be 3·sin(0), and the sine of 0 is just 0. So this ends up just being 0 here. We have this initial velocity in the x-axis, but this still isn't enough. We still only have one variable. We have two unknowns.
This is a situation that may happen often in your physics problems. If you ever get stuck and can't solve using an x-axis equation, then you could always try to solve it with a y-axis equation. You can go to the other axis and try to solve for it there. For instance, we got stuck here in the x-axis. So now we're going to go ahead and try to solve in the y-axis. We're going to pick one of these equations to use, but to do that we need 3 out of 5 variables. We've got the acceleration which is always -9.8. We've got the initial velocity. Remember that this is the initial y velocity at point A, and we know that that's equal to 0. We don't know the final velocity, which would be the y velocity here at the ground.
What about this vertical displacement? The vertical displacement from A to B is just going to be the distance this ball covers from the table down to the ground. We're told that the table's height is 2 meters. So basically, the ball is going to fall 2 meters to the ground. So, this is going to be our Δy, and because this displacement is actually going to be downwards, we're ending up at a lower height than we started from, then this is actually going to be a negative 2 meters because we're always going to use the convention that up and to the right is positive. So any downward displacements are going to be negative.
We've got that Δy and we know that's negative 2, and now we have our 3 out of 5 variables. So we're going to pick the equation that does not have the final velocity. Equation number 3, so we're going to use equation number 3, and we're going to use all of our subscripts. Δy=vayt+12ayt2. This vay turns out to be 0. So now we can simplify and solve for tAB. This Δy is negative 2, and we've got 12·-9.8 times tAB·2. We now solve this equation and tAB ends up being the square root, which is 0.64 seconds. That's the answer to part A.
Let's move on now to part B. We're going to go through the same list of steps again. In part B, we're looking for the horizontal displacement of the ball. We don't have to do the first step again, we already have the path in the x and y. What about this target variable? We're looking for horizontal displacement, this is going to be Δx. The third step is between what interval? We already know we're working from the interval from A to B. So this is going to be Δx+vxtAB here. So now we're just going to go to the UAM equations. Right? Which equation do we use to solve? Well, this is an x-axis variable Δx. So we're going to use the only x-axis equation.
This is going to be Δx=vx·tAB, and this is actually just the same exact equation that we started off with in the first part. But now, we actually know what this time is. So we can actually go ahead and solve because we have both of these variables here. So Δx from A to B is just my 3 meters per second times the 0.64, and that's going to be 1.92 meters. That is the answer to the second part.
Alright. So that's how you solve any one of your projectile motion problems. So I want to make some points here because the type of problem that we just solved was called a horizontal launch. This happens whenever you launch an object horizontally, and there are some special points you should know about this. The first is that the initial velocity is only in the x-axis. What we saw is that when you launch something purely horizontally, all of its velocity is in the x-axis. Your v0x, your initial velocity is just the v0 that you were given. Remember, all objects in projectile motion have zero acceleration in the x-axis. So what that means is that your vx will never change. Whatever this initial velocity is in the x-axis will always be the same exact number throughout the whole entire motion. And the last thing is that because an object is launched horizontally, then the initial component of its velocity in the y-axis v0y is just equal to 0.
Alright? So that's how to solve these kinds of problems. Let me know if you guys have any questions.