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Emily778 Group Title

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

  • one year ago
  • one year ago

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  1. Emily778 Group Title
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    @Nurali

    • one year ago
  2. Emily778 Group Title
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    @Hero

    • one year ago
  3. ranga Group Title
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    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.

    • one year ago
  4. Emily778 Group Title
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    I already told you I have .7

    • one year ago
  5. ranga Group Title
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    First you put that under b. Secondly, that is not k. You have h.

    • one year ago
  6. Emily778 Group Title
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    What do I do next?

    • one year ago
  7. ranga Group Title
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    plug in t = .7 in h(t)=-5t^2+7t and calculate h

    • one year ago
  8. dan815 Group Title
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    lol

    • one year ago
  9. dan815 Group Title
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    there used to be constantly questions about slopes and lines now i see these questions a lot

    • one year ago
  10. dan815 Group Title
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    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

    • one year ago
  11. dan815 Group Title
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    so h'(t) = velocity set h'(t) = 0 and solve

    • one year ago
  12. dan815 Group Title
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    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)

    • one year ago
  13. dan815 Group Title
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    that highest point*, with 0 velocity

    • one year ago
  14. Emily778 Group Title
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    @ranga I got 5.145 for h. Is that right?

    • one year ago
  15. dan815 Group Title
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    oh is this precalc?

    • one year ago
  16. ranga Group Title
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    Try again. Don't forget there is a negative sign in front of t^2.

    • one year ago
  17. ranga Group Title
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    -5t^2+7t -5(0.7)^2 + 7(.7) = ?

    • one year ago
  18. Emily778 Group Title
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    Now you tell me...

    • one year ago
  19. ranga Group Title
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    You can evaluate that using your calculator.

    • one year ago
  20. Emily778 Group Title
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    okay, I got 2.45?

    • one year ago
  21. Emily778 Group Title
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    @ranga

    • one year ago
  22. Emily778 Group Title
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    hello?????????

    • one year ago
  23. ranga Group Title
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    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.

    • one year ago
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