anonymous
  • anonymous
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
Mathematics
schrodinger
  • schrodinger
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anonymous
  • anonymous
anonymous
  • anonymous
ranga
  • ranga
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.

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anonymous
  • anonymous
I already told you I have .7
ranga
  • ranga
First you put that under b. Secondly, that is not k. You have h.
anonymous
  • anonymous
What do I do next?
ranga
  • ranga
plug in t = .7 in h(t)=-5t^2+7t and calculate h
dan815
  • dan815
lol
dan815
  • dan815
there used to be constantly questions about slopes and lines now i see these questions a lot
dan815
  • dan815
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
dan815
  • dan815
so h'(t) = velocity set h'(t) = 0 and solve
dan815
  • dan815
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)
dan815
  • dan815
that highest point*, with 0 velocity
anonymous
  • anonymous
@ranga I got 5.145 for h. Is that right?
dan815
  • dan815
oh is this precalc?
ranga
  • ranga
Try again. Don't forget there is a negative sign in front of t^2.
ranga
  • ranga
-5t^2+7t -5(0.7)^2 + 7(.7) = ?
anonymous
  • anonymous
Now you tell me...
ranga
  • ranga
You can evaluate that using your calculator.
anonymous
  • anonymous
okay, I got 2.45?
anonymous
  • anonymous
anonymous
  • anonymous
hello?????????
ranga
  • ranga
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.

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