Question

when serving in tennis a player tosses the tennis ball vertically in the . The height 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 0.6m above the point of the toss. For how many seconds is the ball in the air between the toss and the serve?
I've already got -7/2x(-5)=7/10-.7

Answer 1

@Nurali

Answer 2

@Hero

Answer 3

For part a)
h(t)=-5t^2+7t
=-5(t^2 - 7/5t)
complete the square of the expression in parenthesis to put the function in the vertex form.
Then comparing it to standard vertex form: h = a(t-h)^2 + k where (h,k) is the vertex.
Identify k and that will be how high the ball goes.

Answer 4

I already told you I have .7

Answer 5

First you put that under b.
Secondly, that is not k. You have h.

Answer 6

What do I do next?

Answer 7

plug in t = .7 in h(t)=-5t^2+7t and calculate h

Answer 8

lol

Answer 9

there used to be constantly questions about slopes and lines
now i see these questions a lot

Answer 10

when you throw a ball up in the air at the highest point its velocity = 0 as it is goes up going up to down, so somwhere in there it has 0 speed and velocity

Answer 11

so h'(t) = velocity
set h'(t) = 0 and solve

Answer 12

that tells you the time it takes to get to highest veloicty then see
what distance you travel given that time substituting that time into h(t)

Answer 13

that highest point*, with 0 velocity

Answer 14

@ranga I got 5.145 for h. Is that right?

Answer 15

oh is this precalc?

Answer 16

Try again. Don't forget there is a negative sign in front of t^2.

Answer 17

-5t^2+7t
-5(0.7)^2 + 7(.7) = ?

Answer 18

Now you tell me...

Answer 19

You can evaluate that using your calculator.

Answer 20

okay, I got 2.45?

Answer 21

@ranga

Answer 22

hello?????????

Answer 23

BTW, the notifications are not working for a couple of days and so people may not be notified when tagged or a reply is posted.

Yes 2.45 meters is correct for a)

For b) substitute h = 0.6 in h(t) = -5t^2+7t and solve for t.
You will get two values for t -- the ball reaches 0.65 meters once while going up and once while coming down. Choose the higher t because they want the value when the ball is coming down.