Pearson+ LogoPearson+ Logo
Start typing, then use the up and down arrows to select an option from the list.

An Unfortunate Landing

Patrick Ford
Was this helpful?
Hey, guys, let's take a look at this problem here. You're gonna take 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. We're gonna figure out the acceleration of the car is 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 flights, 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 we want to do is we want to figure what the acceleration of the car is. So this is just a sorry, not zero. We're actually looking for what this variable is. So because this is ah type of problem in which one object is going to catch up to another one while it's falling, then 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 Siris of steps. First, we're gonna right the position equations for both of them. Then what we do is set them equal to each other so set equal. And the third 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 and step one of our project on motion problems. We know that if this object is gonna take this parabolic path like this. We can split up the X motions in the UAE emotions. So it looks like this. So that's the path of the X and Y. What of our points of interest? Well, really, we're just kicked from the roofs. That's point a, That'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 passed in the X and Y axes. So what is our position equation for the ball Look like? What happens is that we sort of just imagine that this wall here is X initial and we call that zero, then the X of your ball is just gonna be x knots, plus the velocity of the ball Times t plus one half of the acceleration in the X axis times time squared and the ex of your car is gonna be x non plus the car initial T plus one half a car Times T squared. So I'm gonna call the acceleration of a 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 zero, then both of these terms will go away. Now, what about the ball? The ball has an initial velocity, so we won't get rid of that term. But remember that in the X axis, the acceleration is equal to zero. And the only acceleration for the ball happens in the Y axis because of gravity that x downwards. But in the X axis acceleration zero. 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 zero. So really, what these equations actually simplify Teoh is we just have V ball times time and then this is just gonna equal to one half a car. Times t squared. So really, it's just those two equations here, So that's step one. The second thing we're gonna do now is we're actually just going to set them equal to each other. So we have to just set X ball equal to x car. So that means that we're gonna do is we're gonna have the ball times t equals one half of a car. Times t squared, OK? 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 we end up with is we just end up with the 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, the 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 eight. We also know that because this ball is launched purely horizontally, that v a y is just gonna be zero. So that's gonna be, um, so remember. That's always gonna be useful in your horizontal launch problems. So we do know what the 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 toe, 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 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, they're gonna be for the ball. The acceleration is 9.8, the initial velocity in the Y axis va y Because we just said zero. What about the final velocity? That's gonna be the velocity at point B. We've got Delta Y A B. And then ta be Remember, I came to the Y axis because I'm looking for the time. If I could do that, if I could figure that out, you can plug it back into this equation and solve the acceleration. So I need one more variable here. So the final velocity in the Y axis would just be V B y 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 m per second from a 40 m tall building. So what happens is when it falls from a to B. This vertical displacement here is 40 m. However, because this vertical displacement is downwards, then this is just gonna be a negative 40 m. So we know this Delta Y A b is negative 40. And so now we have 3 to 5. So three out of five and we picked the equation. Whoops. And we're going to pick the equation that ignores my final velocity. So if you look through your equations here, that's going to be equation number three. So this is gonna be Delta y A P equals and this is gonna be v A y Times t A B plus one half a y t a b squared, remember. And horizontal launch problems. Um, your V A Y is equal to zero. And so this term always goes away. So you just plug in everything you're gonna have a negative 40 equals one half negative, 9.8 times ta b squared. And if you go ahead and solve for this, what you're gonna get is when you rearrange everything, you're gonna get two times negative, 40 divided by negative, 9.8 and then you're gonna square root that number that's gonna be ta be, and you're gonna get to 0.86 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 the ball, which is equal to eight. That's gonna equal one half times a car, which is what we're looking for times to 86 seconds and you go ahead and work this out. Where you gonna get Is that a car? The acceleration of the car is equal to 5.60 m per second squared. So that's 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. Be so hopefully you guys can see you have to be on the lookout for these kinds of hybrid problems where you really just combining multiple kinds of problems like a catch problem with also a project on motion. But if you just follow the steps for both of them, you'll still get the right answer. That's about this one, guys, let me know if you have any questions.