We should now be familiar with how position and velocity relate to each other using derivatives. And one of the things that you're going to need to do in this course is you're going to need to know how to solve real-world application problems when it comes to velocity or position or just motion in general. We're going to take a look at one of the longer, more complicated types of problems that you're going to see in this course, where you have the real-world situation. So let's take a look. In this problem, we're told the height in meters of a projectile shot vertically upward from a point of 5 meters above ground level with an initial velocity of 20 meters per second looks like this.
So this would be our equation right here. And part a asks us, what is the vertical velocity at t equals 3 seconds? Now we do have a part b and a part c to this problem, but I'm going to take this one step at a time. So starting with part a, we're asked what is the vertical velocity at t equals 3 seconds? Now, the way that you solve these types of problems, if you have a projectile situation, where you are given the equation, this equation is the key to solving the whole problem.
Because this is basically telling you the vertical position. This is the height with respect to time. So if you recognize that this equation of 5t+20t1-4.9t2 is the equation representing the position of your projectile, you can just manipulate this equation in different ways to get all the answers or all the solutions you need in this problem. Now our first piece here asks us to find the vertical velocity at t equals 3 seconds. So if we have a position equation here, how could we find the velocity at one instance in time?
Well, we've talked about how to calculate instantaneous velocity, and we need to recognize that velocity is just the first derivative of position. Now in this case, since we're dealing with a height specifically, a vertical position, we have h of t rather than s of t, but it's really the same thing. So in this case, it's gonna be h prime of t, and that's going to tell you the vertical velocity that you're dealing with. So what I can do here is take the derivative up here. Now the derivative of 5, that's just gonna be 0 because the derivative of any constant is 0.
And we also have 20 t, and the derivative of 20 t is just going to be 20 that t will derive off, and then we're going to have minus 4.9 t squared. Now to find the derivative here, I can use the power rule. Now what I can do is take this 2 here, this squared, I can bring it down to the front front, multiply it by this first number, and then reduce the power by 1. Doing that is going to give me 4.9 times 2, which is 9.8, and we're going to have t to the 2 minus 1, which is just t to the 1 or t. So this right here would be our function.
And to make the velocity function look a little more nice, what I'm going to do is get rid of the 0 because we don't have to write it, and then I'm just gonna write it like this. We're gonna add that our velocity function is 20 minus 9.8 t. So this right here is the velocity function. Now we're specifically asked, what is the velocity at t equals 3 seconds? Well, since we just found our velocity function, I can just take 3 and plug it in.
So we're going to have 20 minus 9.8 times 3. So this is what we get. So all you need to do is plug in these numbers, and if you go ahead and do this on a calculator, you should get negative 9.4. And the units that we're going to have for this is gonna be in meters per second. I know this because I can see that our height here is going to be in meters, and you see the time we're dealing with is in seconds.
A velocity is always gonna be measured in distance per time, so we can see that we have meters per second. So that is going to be the instantaneous velocity at t equals 3 seconds, and that is part a of this problem. But now let's move to part b. Part b asks us, when does the projectile reach its maximum height? Now this is a little bit trickier, because how exactly do we find when the maximum height is going to occur based on this equation that we were given?
Well, to do this, you need to think about the situation. It says that what we do is we start at this position that is 5 meters above the ground. So we're starting 5 meters above the ground, and we have some kind of object. Let's just say it's a ball. And what's gonna happen is this ball is going to be shot directly vertically up in the air, and then eventually it's gonna start coming back down, straight back down.
And we need to figure out exactly when does it reach this maximum height. Well, let's ask ourselves something. What exactly is going to be happening at that position right there? Well, the ball is going or the object of some kind is going to be flying up and up and up, and eventually just for a split second, that object's gonna hold still in the air, then it's gonna reroute and go back down. So we know for that split second it's gonna be holding still as it gets right there.
And because of that, that means that our vertical velocity at that specific instant will be 0. So we now have a definitive place that we know the object is gonna be at its maximum height. So what I can do is take the vertical velocity here and set the whole thing equal to 0, and that's going to give me the time we're at our maximum height. So the way I'm going to do this is I'll take our equation here, 20 minus 9.8, I'll do this in a different color, of 20 minus 9.8 t. I'll set the whole thing equal to 0.
Now all I need to do from here is solve for this t right here. So we're going to take 20 and subtract it on both sides of the equation. That will get the twenties to cancel on the left side, meaning we'll have negative 9.8 tiedade
