 In this video, we provide the solution to question number 14 from practice exam number two for math 1050, in which case we have a toy rocket that's launched up in the air and at some point comes back to the ground, of course. It's initial velocity when it's launched off of the pad there. It's going to be 112 feet per second. So this is our initial velocity. We'll put it v0 to so at a time zero its velocity was such. So the distance s, which we measure in feet of the rocket after t seconds is given by the following equation right here. All right. So honestly, we're just given or handpicked our quadratic equation. How far above the ground will the rocket travel at its highest point? So thinking of this parabolic trajectory of the of the parabola right here, the highest point will be then the vertex of the parabola. So if s is given as 112t minus 16t squared, we then have to figure out what is the h value, right? So that would be the time interval for which you obtain the maximum. That's not the maximum height, but that's where we're going to get to first. So this is the usual formula negative b over 2a. And so b is the linear coefficient. So we get a negative 112 right there. We're going to get two times negative 16, of course. The negative signs cancel out immediately, of course. And then we have to figure out well two times 16 is 32. And how many times does that go into 112? I'm actually going to leave it as a fraction. 16 goes into 112 actually seven times. So this becomes seven halves. So you can leave it as a fraction or if you want a decimal, you could do, you know, 3.5 seconds. All right. So this we should be clear here. This is not the final answer. This is the time. This is the time at the maximum height. This is not the maximum height itself. Why do we care about this? Well, the maximum height, the maximum height would then be obtained at this moment in time. It would be 112 times seven, seven over two minus 16 times seven over two squared. So that's what we're looking for in this situation. All right. So let's then proceed to compute this thing. You know, if you want, again, if you can use the calculator and do this very easy enough, but you know, if you didn't have a calculator at the moment, what would you do? Some things that you can notice, of course, is that there's this fraction of two everywhere. We could factor that out. 112, like I mentioned earlier, 112 is in fact seven times 16. So we actually could factor out two sevens and 16 from everything. So we can factor out those things. So we can factor out a 112 times seven over two. That's common for everything. And that leaves behind then one minus one half, like so. One minus a half, of course, is equal to one half. Two goes into 112. That's going to happen 56 times. We still need the seven. So I'm going to, oh boy, put that seven in there. A little bit of a cheap shot, but we got away with it. So we end up with this 56 times seven over two. Of course, two does go into 56. That's an even number that would leave behind 28, like so. And so we have 28 times seven, which would then give us 196. And we should put the, we should put the units here. This is going to be feet. So the maximum height of the rocket is going to be 196. So the arithmetic there gets a little bit clunky at times. Calculators can help you out with that. That's not really the point of the problem. The point was we have to find the coordinates of the vertex of the parabola for which we found the x-coordinate or the t-coordinate here first. That was the 3.5. And then we put that into the function to find the maximum height, which was 196 feet.