Hey, guys. Let's take a look at this problem here. You're going to kick a ball horizontally off of a building. What happens is a car below on the street is going to accelerate, and then your ball is going to land on the car. We're going to 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 different than some of the problems we've seen before. We've got the roof of the tall building like this. I've got a ball that's on top here, I'm going to kick it with some horizontal velocity. And therefore, it's going to 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 the street right underneath you. Let's say this is a little car like this, and it's going to accelerate from rest and eventually what happens is that their ball, while it's in flight, the car is going to be moving and eventually the car is going to land underneath the ball. Basically, they're just going to catch up to each other. This is a little different than the kinds of problems that we've seen so far. So, what we want to do is we want to figure out what the acceleration of the car is. This is just a, whoops, sorry, not 0. We're actually looking for what this variable is. Since this is a type of problem in which one object is going to catch up to another one while it's falling, then we're going to follow the steps that we use to solve catch problems. Remember that from catch problems, we have a series of steps. First, we're going to write the position equations for both of them. Then what we do is set them equal to each other. Lastly, the third step is we solve for time.

We'll start with the position equations for both of these objects. Before we do that, we're going to draw the paths on the x and y axes in step 1 of our projectile motion problems. Knowing that if this object is going to take this parabolic path like this, we can split up the x motions and the y motions, so it looks like this. What are our points of interest? Really, we're just kicked from the roof so that's point A, it's our initial. 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 on the X and Y axes. What does our position equation for the ball look like? If we imagine that this wall here is x initial and we call that 0, then the x of your ball is just going to be x₀ + velocity of the ball times t + one-half of the acceleration on the x-axis times time squared. And the x of your car is going to be x₀ + v_car initial t + one-half a_car times t squared. I'm going to call the acceleration of the car a_car, and that's what we're looking for here. That's the target variable.

Now that we've written out the position equations, let's see if we can simplify them. If we 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 on the x-axis, the acceleration is equal to 0. The only acceleration for the ball happens on the y-axis because of gravity acting downward. But on 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 going to be 0. So really, these equations actually simplify to just v_ball times time, and this is going to equal 1/2 a_car times t squared. So it's just those two equations here.

Next, we're just going to set them equal to each other. We have to set x_ball equal to x_car. So what that means is we're going to have v_ball times t equals one-half of a_car times t squared. Now the third step is we're just going to solve for the time. Notice here how we can cancel out a factor of time from both sides of the equation. What we end up with is v_ball = 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, v_ball is just the initial velocity given to the ball, which is really just the velocity in the x-axis, is just 8 m/s.

If we're trying to figure out the acceleration, and I don't know the time, it's the time that it takes for the ball to actually travel through the air and then finally hit the ground. Just like any projectile motion problem, I've gotten stuck on the x-axis trying to solve for time. So what do I do? I go to the y-axis. In the y-axis, there's only one thing moving. It's just the ball going from the roof down to the ground. So all of these y-axis variables are for the ball. The acceleration is 9.8 m/s². The initial velocity in the y-axis, v_a y, is 0. What about the final velocity? That's going to be the velocity at point B. We've got delta y from a to b and then t from a to b. I'm looking for the time. We know the vertical displacement from a to b, which when the ball is kicked at 8 m/s from a 40-meter tall building, this vertical displacement here is 40 meters. However, because this vertical displacement is downwards, then this is just going to be -40 meters.

We now have 3 out of 5 variables, and we pick the equation that ignores my final velocity. Delta y_a_b equals v_a_y times t_a_b plus one-half a_y t_a_b squared. In horizontal launch problems, your v_a_y is equal to 0, and so this term always goes away. Plugging in everything, -40 equals 1/2 (-9.8) times t_a_b². Solving for this, you get 2 * (-40) / (-9.8), and then square rooting that number, you get t_a_b and you get 2.86 seconds. Now that we finally have this time, we can plug it back into our equation and solve for the acceleration. v_ball, which is 8, is going to equal 1/2 times a_car, which is what we're looking for, times 2.86 seconds. Working this out, you get that a_car, the acceleration of the car, is 5.60 m/s². That's how fast the car has to accelerate in order to land 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.