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anonymous

  • one year ago

The function H(t) = -16t2 + vt + s shows the height H(t), in feet, of a projectile launched vertically from s feet above the ground after t seconds. The initial speed of the projectile is v feet per second. Part A: The projectile was launched from a height of 96 feet with an initial velocity of 80 feet per second. Create an equation to find the time taken by the projectile to fall on the ground. Part B: What is the maximum height that the projectile will reach? Show your work. Part C: Another object moves in the air along the path of g(t) = 31 + 32.2t where g(t) is the height, in fe

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  1. anonymous
    • one year ago
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    the object from the ground at time t seconds. Use a table to find the approximate solution to the equation H(t) = g(t), and explain what the solution represents in the context of the problem? [Use the function H(t) obtained in Part A, and estimate using integer values] (4 points) Part D: Do H(t) and g(t) intersect when the projectile is going up or down, and how do you know?

  2. anonymous
    • one year ago
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    @perl @perl

  3. perl
    • one year ago
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    Which part would you like help with?

  4. anonymous
    • one year ago
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    all of it.....

  5. perl
    • one year ago
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    Ok we can start with part a)

  6. anonymous
    • one year ago
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    maybe once i get started i MIGHT be able to do it thanks so much

  7. anonymous
    • one year ago
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    H(t) = -16t^2 + 80t + 96

  8. anonymous
    • one year ago
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    i think ^^

  9. perl
    • one year ago
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    We are given that \( H(t) = -16t^2 + vt + s\) shows the height H(t), in feet, of a projectile launched vertically from s feet above the ground after t seconds. The initial speed of the projectile is v feet per second. Part A) The projectile was launched from a height of 96 feet with an initial velocity of 80 feet per second. That means s = 96 and v = 80

  10. perl
    • one year ago
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    yes that's correct :)

  11. anonymous
    • one year ago
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    ok so my equation becomes : H(t) = -16t^2 + 80t + 96

  12. anonymous
    • one year ago
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    ok so thats part A right?

  13. anonymous
    • one year ago
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    now part b i have to find the maximum

  14. anonymous
    • one year ago
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    so the axis of symmetry is 2.5?

  15. perl
    • one year ago
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    the maximum or minimum of a parabola occurs at \(\large t= \frac { -b}{2a} \)

  16. anonymous
    • one year ago
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    ok so I got a maximum of 196

  17. perl
    • one year ago
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    how plug in that time t = 2.5 to find the maximum height

  18. anonymous
    • one year ago
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    196 :)

  19. perl
    • one year ago
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    correct

  20. anonymous
    • one year ago
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    ok so now part c... now i really dont know where to start

  21. perl
    • one year ago
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    Part C) the directions are not complete . Can you check

  22. anonymous
    • one year ago
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    yes hang on one sec please

  23. anonymous
    • one year ago
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    Part C: Part C: Another object moves in the air along the path of g(t) = 31 + 32.2t where g(t) is the height, in feet, of the object from the ground at time t seconds.

  24. anonymous
    • one year ago
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    Use a table to find the approximate solution to the equation H(t) = g(t), and explain what the solution represents in the context of the problem? [Use the function H(t) obtained in Part A, and estimate using integer values]

  25. perl
    • one year ago
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    We want to find when \( H(t) = g(t) \) which would ordinarily mean solve the equation \( -16t^2 + 80t + 96= 31 + 32.2t \) but instead of solving for t , here they ask us to make tables for the left side and right side separately, and estimate when they are closest

  26. perl
    • one year ago
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    |dw:1435527312454:dw|

  27. anonymous
    • one year ago
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    ok so I would solve each equation both H(t) and g(t) by plugging in the numbers 1-5?

  28. perl
    • one year ago
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    we can find y values for both H(t) and g(t) and then estimate where they are closest

  29. anonymous
    • one year ago
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    ok so we would do that by plugging in 0 right?

  30. perl
    • one year ago
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    yes

  31. anonymous
    • one year ago
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    ok so g(t) = 31 + 32.2t , g(t) = 31 + 32.2(0) , g(t) = 31 + 0 , g(t) = 31 H(t) = -16t^2 + 80t + 96 , H(t) = -16(0)^2 + 80(0) + 96 , H(t) = 0 + 0 + 96 , H(t) = 96

  32. anonymous
    • one year ago
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    so the y-intercept for H(t) = -16t^2 + 80t + 96 is 96 and the y-intercept for g(t) = 31 + 32.2t is 31

  33. anonymous
    • one year ago
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    @perl

  34. anonymous
    • one year ago
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    can you help me with part d @perl

  35. anonymous
    • one year ago
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    equation 1: (4,160) equation 2: (4, 159.8) what does it represent in the context of the problem?

  36. perl
    • one year ago
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    |dw:1435528016137:dw|

  37. perl
    • one year ago
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    |dw:1435528443369:dw|

  38. anonymous
    • one year ago
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    they almost intersect when x = 4 what is the meaning in the context of the problem?

  39. anonymous
    • one year ago
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    @perl

  40. perl
    • one year ago
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    at time t = 4 seconds the paths are closest to each other

  41. perl
    • one year ago
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    the projectiles are closest

  42. anonymous
    • one year ago
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    ok thanks! what about part d?

  43. anonymous
    • one year ago
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    when the projectiles re going down because H(t) = -16t^2 +80t +96 is decreasing just as the other equation is increasing?

  44. anonymous
    • one year ago
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    @perl

  45. anonymous
    • one year ago
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    @perl is that correct??

  46. perl
    • one year ago
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    when the projectile h(t) is coming down , yes

  47. perl
    • one year ago
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    the projectile g(t) is going up the whole time

  48. anonymous
    • one year ago
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    ok thanks so much !!!!!!

  49. perl
    • one year ago
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    |dw:1435528915392:dw|

  50. perl
    • one year ago
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    rough picture :)

  51. anonymous
    • one year ago
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    THANK YOU SO MUCH!

  52. perl
    • one year ago
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    Your welcome

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