anonymous
  • anonymous
A projectile is thrown upward so that its distance above the ground after t seconds is given by the function h(t) = -16t2 + 640t. After how many seconds does the projectile take to reach its maximum height?
Mathematics
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SOLVED
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schrodinger
  • schrodinger
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Smartanne
  • Smartanne
I can help
anonymous
  • anonymous
thank you
anonymous
  • anonymous
i can help

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More answers

tkhunny
  • tkhunny
Have you seen this? \(-\dfrac{b}{2a}\)
anonymous
  • anonymous
It will reach maximum height when h'(t) = -32t + 640 = 0 -32t = -640 t = -640/-32 = 20 seconds
anonymous
  • anonymous
thank you, could you explain how you did that? @tdunmyer
anonymous
  • anonymous
i asked my cousin the question and he told me to put what he said he's a math teacher
tkhunny
  • tkhunny
We should probably first determine if the Calculus can be used. If this is Algebra 2 or Trig, we need a different solution.
anonymous
  • anonymous
this is pre calculus @tkhunny
anonymous
  • anonymous
oh okay. well thank you! @tdunmyer
tkhunny
  • tkhunny
Okay, now answer my first question. Have you seen, \(-\dfrac{b}{2a}\)? If not, we'll have to derive it.
anonymous
  • anonymous
it looks familiar, but I don't really know what it is @tkhunny
tkhunny
  • tkhunny
It's a neat little idea that can be derived several ways. h(t) = -16t2 + 640t a = -16 b = 640 -b/(2a) = -640 / (2*(-16)) = -640/(-32) = 20 <== Same answer as the calculus version. It tells you the value of 't' for which the height is maximum. It does NOT tell you the maximum height. You must evaluate the function to see that. You can find -b/2a by 1) Completing the Square 2) Finding the average of the two quadratic solutions. 3) Using a derivative from calculus. 4) There may be other ways. Here's #2 If we use the quadratic formula, we get for f(t) = 0 \(t = \dfrac{-b\pm\sqrt{Stuff}}{2a}\). Find the average of those 2 \(\dfrac{\dfrac{-b - \sqrt{Stuff}}{2a} + \dfrac{-b + \sqrt{Stuff}}{2a}}{2} = \dfrac{\dfrac{-2b}{2a}}{2} = -\dfrac{b}{2a}\). If you can find where the projectile is on the ground, there should be two places, the average of those two places is the where the maximum occurs.

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