Hey guys, let's work this problem out together here. So we're gonna launch ball directly upwards from the ground. Let me go ahead and start drawing that. So we've got the ground level like this, y equals zero. I've got a ball that I'm gonna launch upwards with some initial speed. That's actually the first part of the problem. I want to calculate that launch speed so this V equals something that I don't know. What I do know about this problem is that at some later time the ball is still going upwards, I'm told that here the ball has a speed of 20 m per second and the height is equal to 30 and because it's still moving upwards at some height you can actually still going even higher than this. So it's going to continue going upwards until finally reaches its maximum height. Here, we know that the speed is going to be zero and this is going to be why max that's actually be the part B of our problem. So party, we're going to calculate this launch speed, part B. We're going to calculate this wine max. So this is our diagram here. We want to use energy conservation because we have changing heights and changing speeds. So what happens is we're gonna write our energy conservation equation. But what is going to be our initial and our final, we actually have three different points here. So there's a couple of accommodations for initial and final. I can go from here. Here we go from here to here. So what happens to these problems is that some problems will give you more than two points of information. Some points aren't just going to give you as simple as an initial and final to give you some other information. So what happens is I'm going to label these as a B and C. Similar to what we did with projectile motion. So A is when we're at the ground, B is when we know that we're 20 m/s, 30, 30 m and then see is going to be the maximum heights. So we're gonna have to write an energy conservation equation and the two points that we're gonna pick our initial and final should be given and the targets the given should be the one that you know everything about. The target variables should be the target of interval or the point of interest should be the thing that you actually are looking for. So for example, we're looking for the launch speed, which is vous we want to pick A is one of our points of interest and then between B and C, the one that makes more sense is be because we know everything about it. If we were to try to pick point C, is there another point of interest like from initial to final? We would actually sort of get stuck because we would have two unknowns in that equation why max and V. A. So we actually wouldn't be able to solve it that way. All right. So for part A. Here, I'm going to write my conservation of energy from point A. To B. So I'm gonna do the kinetic initial which is K. A. Plus the potential initial is equal to the kinetic final plus the potential final. We're just using a zombies because that's my initial and final here. So let's go ahead and eliminate expand all of our terms. We definitely have some kinetic energy. That's what we're looking for, what we're looking for the speed. We don't have any gravitational potential because y is equal to zero. This is our sort of floor level or ground level. So we're gonna just set gravitational potential to be zero there and just we can get rid of that term and then there's definitely gonna be some kinetic and potential when we get to point B. Because we have some speed and some height. So let's go out and expand each of the terms here. We're going to have one half M V A squared. That's our target variable equals one half M V B squared plus mg times Y B. All right. So I'm going to do here is as expected, are masses are going to cancel from all the sides of the equation. And that actually is good for us because we didn't know that what the mass of the ball was, but it turns out doesn't matter. And we can also multiply this equation by two because we have a bunch of fractions in here. So we're gonna get V A squared equals V. B squared plus. And then when you multiply it by two, you're gonna have to insert a factor of two here. This is gonna be 2G. Why be? So now we're just gonna take the square roots and start plugging in numbers. So via is gonna be the square roots, This is going to be 20 squared plus two times 9.8 times the height at point B. Which we know is 30. If you go ahead and work this out, you're gonna get 31.4 m per second. So that's the answer here. At the sort of launch speed or launch speed was actually equal to 31.4 m per second. And then it is going to slow down when it gets to point B. It's gonna only be traveling at 20 m per second and continue one until it reaches a point C at which it stops. All right. So that's the answer to part A. All right. So let's take a look at part B now and now we want to calculate the maximum height. So that's going to be this guy over here. Now. Again, we want to pick an interval in which we know something about one interval and we're looking for Y max and it turns out we can actually use either the interval from A to C or the interval from B to C. Notice how we have all this information now, so we know that this Y. A. Is equal to zero. So we know everything about both of these points of interest here. It actually doesn't matter which one we pick. So just to make things a little bit simpler, I'm going to choose from A to C just because we know that one of the terms is going to cancel out. So let's go ahead and write this right. We've got K. A. Plus you A. Equals K. C. Plus you see, so we know there's no potential energy initial. And now what happens is the point. See there's gonna be no kinetic energy final. And so what happens is we're gonna get one half M V. A squared equals and this is going to be M. G. Y. C. Or we can call this actually why max here as well. So that's our target variable are masses will cancel and we actually know what this via is now. So we can go ahead and solve. Let's see, we're just gonna get um this Y max here is equal to this is gonna be V. A. Which is the 31.4. So we just figured out, divided by two times 9.8. I'm just doing one half of that velocity, divided by the G once we move it over. And so what you end up getting here is a wimax of 50.3 m. So that's the maximum heights. All right. So this thing goes 50.3 m. Final makes sense because here we were at 30 m, but we still had some speed. So we should expect this to be a higher number. All right. So that's it for this one. Guys, let me know if you have any questions.