A projectile is fired vertically upward and has a position giv...

Question

A projectile is fired vertically upward and has a position given by s(t)=−16t^2+128t+192, for 0≤t≤9.

a. Graph the position function, for 0≤t≤9.

Accepted Answer

Welcome back, everyone. In this problem, a ball is thrown directly upwards, giving its height as a function of time T by H of T equals -5T 2 + 40 T + 50 for values of T between 0 and 8. Plot the function for T between 0 and 8. Here we have for our answer choices possible plots for our function. A, B, C, and D, OK. And then here I also have a graph that we're going to use to do our plot. So let's see if we can go ahead and plot our function 4. Let me rewrite our function here H of T. Equal to negative 5T 2 plus 40T + 50, OK, for T between 0 and 8. Now, to plot the function H of T, we will first calculate key points on the graphs, such as the vertex and the Y intercept, and then we're going to plot additional points within the given interval. So let's start first start by finding the Y intercept, OK? And know to find the Y intercept. This is the point at which the value of T is going to be equal to 0, OK? So the Y intercept. Is going to be H of 0, OK. Evaluating H at T equals 0, and no, that means we're going to substitute 0 wherever we see T in our equation. So that's -5 multiplied by 0 squared plus 40 multiplied by 0 + 50, OK, which is going to be 0 + 0 + 50, which equals 50, OK? So that means the Y intercept is at 050. So we can keep that in mind. Next, let's try to find the vertex, OK? Now to find a vertex. We know that the vertex of a parabola. Of a quadratic function in the form AX2 + BX + C is going to be negative B divided by 2A for the X coordinate and H of whatever that value is, so H of negative B divided by 2A4 or Y coordinate. So first, we need to evaluate this expression and then use it to figure out our X and Y coordinate. Now, as we said, this is for a parabola, OK, which is in the general form of AX2 + BX + C. No, for a function H of X, OK, for H of T sorry, H of T rather, A is equal to -5, OK. That's the coefficient of the squared term. B is equal to 40 the coefficient of the linear term and the constant equals 50. Therefore, OK, the X coordinate. Of the vertex Is going to be equal to -40 divided by 2 multiplied by -5. That's -40 divided by -10 which equals 4. And since neb divided by 2 A equals 4, then this means the Y coordinate. Is going to be H of 4. And that is going to be equal to -5 multiplied by 42 + 4 multiplied by 5 + 50. No, sorry, 4 square is 16, -5 multiplied by 16 equals -80, 40 multiplied by 5 equals 200. Sorry, not too hot. 40 multiplied by 4 rather this should have been 4. My apologies here. And 40 multiplied by 4 equals 160 and 80 + 160 + 50 equals 130. So what does this tell us? This means the vertex of our of our graph. is at 4130, OK. That's another Coordinate that we'll have to keep in mind. Now, finally, let's plot some additional points. And since we've already found it for uh H of 0 and H of 4, we can use some even values in between, OK? So, for example, we could find The value for a T equals 2, and shows not random. So that's gonna be H of 2, which is -5. Multiplied by 2 squared. Plus 40 multiplied by 2 + 50. Which equals -20 + 80 + 50, which equals 110, which tells us then that 2 110 is also going to be on our graph. And we could use another point, say, for T equals 6. And now for T equals 6, H + 6 is going to be equal to -5 multiplied by 6 squared, plus 40 multiplied by 6 + 50. That's -180 + 240 + 50, which equals 110 as well, OK? So, 6, 110 is going to also be on our graph. And at this point, I think we have enough points, uh, we, we have enough coordinates to plot our graph. So we have the Y intercept 050, the vertex 4130, and the 2 points 21,010 and 6 10,010. So we're going to plot our points and then connect the points with a smooth curve to represent the parameter. I'm going to try to draw a curve as smooth as I possibly can. Now you notice for the graph that I have here, OK. Notice that on our X-axis, each increment goes up by 10 units, OK. Each unit is worth 10, sorry, on the Y axis, each unit is worth 10, and on our X-axis, each unit is worth 0.2. So now, if we go ahead and plot our points, they are 050. OK. 2 110, be right here. Uh 4 130, OK, which would be right here, and then 6, 110. We have these points, so now we can go ahead and plot. So our curve is probably going to look something like this, OK? Let me just make sure that I have it properly here. It's gonna look something like this. Oh, that was so close. Let me just draw one more just to make sure that I get it as close as possible. So yes, it would look something like this. So this would be our curve or or function of H of T. Now if we compare our result with the rest of answer choices, OK, then we can tell that D is the correct answer. This is the graph of D. As a matter of fact, if we go back to D here, we can tell that 050, 2 110, 4 130, and again 26 110 are on this curve. So D is the correct answer. Thanks a lot for watching everyone. I hope this video helped.

A projectile is fired vertically upward and has a position given by s(t)=−16t^2+128t+192, for 0≤t≤9.

a. Graph the position function, for 0≤t≤9.

Key Concepts

Quadratic Functions

Graphing Techniques

Projectile Motion

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