5. Projectile Motion
Intro to Projectile Motion: Horizontal Launch
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Hey, guys. Let's take a look at this problem here. You're gonna kick a ball horizontally off of the building. And what happens is a car on underneath you or below on the street is gonna accelerate, and then your ball is gonna land on the car. And we're gonna figure out what the acceleration of the car is. So before we go into anything, let's just go ahead and draw this out because this looks a little bit different than some of the problems we've seen before. So we've got the roof of the tall building like this. So I've got a ball that's on top here, I'm gonna kick it with some horizontal velocity. And so therefore, it's gonna take this sideways or parabolic path as it goes down. Now at the same time, there's a car that's at the bottom right on on the street right underneath you. So let's say this is a little car like this, and it's gonna accelerate from rest and eventually what happens is that their ball, while it's in flight, the car is gonna be moving and eventually the car is gonna land, underneath the ball. Basically they're just gonna catch up to each other. So this is a little bit different than the kinds of problems that we've seen so far. So what we wanna do is we wanna figure out what the acceleration of the car is. So this is just a, whoops, sorry, not 0. We're actually looking for what this variable is. So because this is a type of problem in which one object is going to catch up to another one while it's falling, then what we're gonna do is we're gonna follow the steps that we use to solve catch problems. And remember that from catch problems, we have a series of steps. First, we're gonna write the position equations for both of them. Then what we do is set them equal to each other. So set equal. And then 3rd step is we solve for time. Now we're basically gonna do those exact steps over here. Let's just go ahead and start out with the position equations for both of these objects. So first, before we do that, we're gonna draw the paths in the x and y axis in step 1 of our projectile motion problems. We know that if this object is gonna take this parabolic path like this, we can split up the x motions and the y motions, so it looks like this. So that's the path in the x and y. What are our points of interest? Well, really we're just kicked from the roof so that's point a, it's our initial. And then nothing happens in between and then it finally just lands on the car on the ground at point b. So those are our paths in the X and Y axis. So what does our position equation for the ball look like? What happens is if we sort of just imagine that this wall here is x initial and we call that 0, then the x of your ball is just gonna be x naught plus velocity of the ball times t, plus one half of the acceleration in the x axis times time squared. And the x of your car is gonna be x naught plus v car initial t, plus one half a car times t squared. So I'm gonna call the acceleration of car a car, and that's actually what we're gonna be looking for here. That's the target variable. So we're looking for what is a car. Okay. So now that we've written out the position equations, let's see if we can simplify them. Well, if we just call the initial position 0, then both of these terms will go away. Now what about the ball? The ball has some initial velocity, so we won't get rid of that term. But remember that in the x axis, the acceleration is equal to 0. And the only acceleration for the ball happens in the y axis because of gravity that acts downwards. But in the x axis, the acceleration is 0, so that term also goes away. And then for the car, what happens is we're told that the car accelerates from rest, which means that the v initial of the car is gonna be 0. So really what these equations actually simplify to is we just have v ball times time, and then this is just gonna equal to 1 half a car times t squared. So really, it's just those two equations here. So that's step 1. The second thing we're gonna do now is we're actually just gonna set them equal to each other. So we have to set x ball equal to x car. So that means that we're gonna do is we're gonna have v ball times t equals one half of a car times t squared. Okay? And so now the third step is we're just gonna solve for the time. So notice here how we can actually cancel out a factor of time for both sides of the equation here. And then what we end up with is we just end up with v ball equals one half of a car times one factor of t. Remember, in this equation or in this problem, we're looking for a car. Now what is v ball? Well, v ball is just the initial velocity that's given to the ball, and we know that v ball, which is really just the velocity in the x axis, is just 8. We also know that because this ball is launched purely horizontally, that v a y is just gonna be 0. So that's gonna be, so remember, that's always gonna be useful in your horizontal launch problems. So So we do know what v ball is. Unfortunately, if we're trying to figure out the acceleration, I don't know what the time is. That's the time that it takes for the ball to actually travel through the air and then finally hit the ground. And so just like any projectile motion problem here, I've gotten stuck in the x axis trying to solve for time. So what do I do? I just go to the y axis. And in the y axis, there's only one thing that's moving. It's just the ball that's going from the roof down to the ground. So all of these y axis variables are gonna be for the ball. The acceleration is 9.8. The initial velocity in the y axis, v a y, is we just said 0. What about the final velocity? That's gonna be the velocity at point b. And we've got delta y a b and then t a b. Remember, I came to the y axis because I'm looking for the time. If I can do that, if I can figure that out, I can plug it back into this equation and solve for the acceleration. So I need one more variable here. So the, final velocity in the y axis would just be Vvy like this. I don't know what that is. What about the vertical displacement? What about delta y from a to b? Well, if I look at my diagram here, and if I look at my problem, what I'm told is that the ball is kicked at 8 meters per second from a 40 meter tall building. So what happens is when it falls from a to b, this vertical displacement here is 40 meters. However, because this vertical displacement is downwards, then we this is just gonna be a negative 40 meters. So we know this delta y a b is negative 40, and so now we have 3 to 5. So 3 out of 5, and we pick the equation. Oops. And we're going to pick the equation that ignores my final velocity. So if you look at through your equations here, that's going to be equation number 3. So this is gonna be delta yab equals, and this is gonna be, vaytimestabplusone half a ytab squared. Now remember in horizontal, launch problems, your v a y is equal to 0, and so this term always goes away. So we just plug in everything. The net negative 40 equals 1 half negative 9.8 times TAB squared. And if you go ahead and solve for this, what you're gonna get is when you rearrange everything, you're gonna get 2 times negative 40 divided by negative 9.8, and then you're gonna square root that number, that's gonna be TAB and you're gonna get 2.86 seconds. So now that we finally have this time here, we can plug it back into our equation and solve for the acceleration. So we've got v ball which is equal to 8, that's gonna equal 1 half times, a car, which is what we're looking for, times 2.86 seconds. And we go ahead and work this out, what you're gonna get is that a car, the acceleration of the car is equal to 5.60 meters per second squared. So that's how how fast the car has to accelerate in order to basically fall right underneath the ball as it hits the ground. So that is answer choice b. So hopefully you guys can see that you have to be on the lookout for these kinds of hybrid problems where you're really just combining multiple kinds of problems like a catch problem with also a projectile motion. But if you just follow the steps for both of them, you'll still get the right answer. That's it for this one, guys. Let me know if you have any questions.
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