A 250 g ball is launched with a speed of 35 m/s at a 30° angle. A strong headwind exerts a constant horizontal drag force on the ball. What is the magnitude of the drag force if the wind reduces the ball's travel distance by 20%?

Video transcript

Hey, everyone. So this problem is dealing with projectile motion. Let's see what they're asking us. We have a rugby player kicking a 400 grand ball with an initial speed 25 m per second. The initial velocity vector makes an angle of 40 degrees with the horizontal, the wind blowing in the stadium applies a constant horizontal drag force of a given magnitude F D to the moving ball. As a result, the ball's displacement is divided by two compared to the ball's displacement if there was no drag force. And then they're asking us to calculate what that F D this drag force is okay. So the first thing we can do is recall our cinematics equations. And so for this problem, we're going to be utilizing delta X equals one half A X T squared plus the not X T and B Y equals V, not Y plus A Y T. And we really have two different scenarios here. We have a scenario where the ball has a certain displacement if there is no drag force. And then, so we'll call that scenario one or scenario with no drag force. And then we have a separate scenario where we do have that drag force and therefore the displacement is smaller and it's actually given to us that it is twice as small. So the first thing we're going to do is look at what this no drag scenario looks like. So we have the tracking of the projectile of this ball that is kicked. And we know that at the inflection point, our velocity V Y is going to be zero, right? So at that, at that instant, right before it starts to come down, it's for it's vertical velocity component is zero. And then we also know that because we are working in a symmetrical flight path, the time that it takes for the ball to go up is going to be the same amount of time as it takes for the ball to go down. And so we can call both of those T up and this total time is gonna be t another way we can write that is just to T up equals T. Alright. So back to our displacement here, we've got delta X, we'll call no drag sub N D. There is no acceleration term because there are no forces acting on the ball in the X direction. And so the only uh so that turn cancels out. So the only term working here for delta X is going to be V not X T. And we can recall that the X component of that velocity vector is given by V naught times post sine of theta times T. And for V Y, we have V Y equals V, not Y. It's gonna be very similar where we can recall that V naught times this sign of data. And then in this case, our acceleration in the Y direction is just gravity, there are no other forces acting on it. So we're going to rewrite that as minus G because gravity is acting in negative or downward direction. T we also know that at the time T equals T up R V Y is zero. So that looks like zero equals B, not sine theta minus G T up. And remembering that to tee up equals T, we can simplify this to solve for T and that gives us to be not sign data over G. So we are going to do the same thing for our second scenario where we do have the drag force, the time of this ball's flight is going to be the same in both scenarios. So we can also solve for, we can also solve these kids matics equations for time in the second scenario. And then use that to help us find our drag force. Alright. So I'm gonna rewrite the cinematics equations here. So this is scenario two with drag. So I'm gonna write that as delta X sub D equals one half A X T squared plus being not X T. And so this delta X D From the problem we are told that Delta XD is equal to 1/2 delta X N D. And we can also recall that when we have just kind of draw a free body diagram over here on the side. And so when we have our ball, the only force acting on it in the X direction is our drag force, and then we'll have weight acting on it in the Y direction. And so from Newton's second law, we can take the sum of the forces in the X direction equal to mass times acceleration. And so that will be minus F D equals A or solving for A, we have minus F D over. And so now when we go back to this first cinematics equation for our drag scenario, we can start substituting in other equations, other things that we've already worked out from the problem and putting things in terms so that we can solve for this F sub D R drag force. So the first thing we're gonna do take this delta X sub D and recall that that's one half times delta X have no drive, which we've identified here is the not cosign theta T Write that down here equals 1/ A X we've solved for here is minus F D over em T squared plus and then V not X already signed, we already solved out or simplified that out to be not chosen data. So I'm gonna do that here again to T All right. And so now we can further simplify this, we will subtract this V nought cosine theta T term from both sides. So we'll have minus one half V, not cosign theta T equals minus one half F D over M T square. And now we can start to cancel things out. Simplify and saul bore T. So one T left here that looks like T equals V, not cosign theta em over sub D as I said before, we know that the times are the same. And so we can set these two equations equal to each other here and here that gets us us. Third equation of the not co sign theta M over F D equals to be not signed data over G are the notes cancel and we can rearrange this to get our drag force by itself. So that looks like co sign theta times M G over to sign data. And from here we are finally ready to plug in some numbers that we've gotten from our problem and solve for our force. So we recall the problem text gave us a mass of 400 brands. So working in Standard units, I'm just gonna rewrite that as 0.4 kg. Data from the problem was 40 degrees. And then G gravity is a constant that we can recall is 9.8 m per second square. And so now we're ready to plug all of that in. So we've got the coastline of 40 times 0. kg times 9.8 m per second squared, All over two times a sign of baton To our calculators. And we are left with a force drag force of 2. newtons. So that is the answer to this problem. We go back up to our multiple choice and we can see the bad alliance of choice. See that's all we have for this problem. We'll see you in the next video.