Question

When serving in tennis, a player tosses the tennis ball vertically in the air. The height of h of the ball after t seconds is given by the quadratic function h(t) = -5t^2 + 7t (the height is measured in meters from the point of the toss).

a. How high in the air does the ball go?
b. Assume that the player hits the ball on its way down when it's 0.6 m above the point of the toss. For how many seconds is the ball in the air between the toss and the serve?

Answer 1

I want someone to explain this to me rather than giving me the answer.

Answer 2

Do you know what the graph of this function looks like? You can graph it and it sort of looks like the path of something flying through the air. That helps you visualize what you need to find.

Answer 3

Right. I'm not exactly sure how to solve the function though. I have to plug something in somewhere don't I? And if I do, I'm not sure how to get there.

Answer 4

For part a, you want to know how high the ball goes.

Answer 5

Do you know how to do that?

Answer 6

No. That's what I need help with.

Answer 7

There are two ways to do it, the looking at parabolas way and the calculus way. Which one is "right" depends on your course, I guess. What course are you in?

Answer 8

It's Algebra 2, so looking at parabolas would make more sense.. or at least to me I think it would.

Answer 9

Ok. The vertex of a parabola is \(-\frac{b}{2a}\)
You should probably just remember that formula; it comes in handy.

Answer 10

Got it. Now this is where I get confused. I'm not sure what the next step is.

Answer 11

You have the x coordinate. So just find the y coordinate and that's your height.

Answer 12

Yeah... not sure what the x coordinate is.

Answer 13

I just told you. :D
\[-\frac{b}{2a}\]
That's the x coordinate of the vertex of the parabola.

Answer 14

Oh! Okay.. so where do I plug it in though?

Answer 15

Where do you plug what in?

Answer 16

Don't I have to plug \[-\frac{ b }{ 2a }\] into the quadratic function?

Answer 17

Well, yes, but you can evaluate it first.
The quadratic equation is in the form \(f(x) = ax^2 + bx + c\).
That's where you're getting the a and b from.

In the case of \(-5x^2 +7t\) it's
\[-\frac{7}{2(-5)}\]

Answer 18

Oops, \(-5x^2 + 7x\). Or \(-5t^2 + 7t\). Just one variable.

Answer 19

Okay. Then solving this will give me the answer to a correct?

Answer 20

That will give you the x-coordinate of the vertex of the parabola.
Read the problem. Is that what you need to find?

Answer 21

Well I know that h is the height and t is seconds and that I need to find how high the ball goes. So if I graph to function and I have the x-coordinate that will give me the answer to a? Is that what you mean?

Answer 22

No.

The function is height as a function of time. That is, on that graph, the x-axis represents time, and the y-axis represents height. Therefore, finding the x-coordinate of the vertex means that you will find the time the ball is at its highest. You want the highest that the ball goes.

Answer 23

Okay. I guess some of this is just confusing.

Answer 24

It can be. But that's why I'm trying to explain it. :D

Answer 25

Do you think you could go through the steps again?

Answer 26

I'm not sure what retyping the same stuff would do. Scroll up. Then ask a more specific question.

Answer 27

So I get that \[-\frac{ 7 }{ 2(-5) }\] is going to be \[\frac{ 7 }{ 10 }\] Now I'm just not sure what to do after that.

Answer 28

You're given height as a function of time. You have the time you want. You just need to find the height.

So, you just need to plug 0.7 into your original function, \(-5t^2+7t\). That tells you the height at 0.7 sec, which also happens to be the highest point of the parabola.

Answer 29

Okay so I got 2.45

Answer 30

Ok. So that's the highest height of the ball.

Answer 31

Cool. so for b I would just plug 0.6 into the height then?

Answer 32

Right, and then solve for time.

Answer 33

Awesome! Thank you so much! :)

Answer 34

Glad to help. :D