anonymous
  • anonymous
An object is launched upward at 45 ft/sec from a platform that is 40 feet high. What is the objects maximum height if the equation of height (h) in terms of time (t) of the object is given by h(t) = -16t2 + 45t + 40? 65 feet 72 feet 80 feet 93 feet
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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chestercat
  • chestercat
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Mehek14
  • Mehek14
what did you get?
anonymous
  • anonymous
okay so, ik i can take a 4 out of all of them
welshfella
  • welshfella
maximum height is when velocity = 0

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Mehek14
  • Mehek14
use the same formula as last time
anonymous
  • anonymous
okay -45/2(-16)
Mehek14
  • Mehek14
yes and then?
anonymous
  • anonymous
-45/-32 1.4
Mehek14
  • Mehek14
it can actually be rounded to 1.41
Mehek14
  • Mehek14
now plug 1.41 into the equation
anonymous
  • anonymous
okay then plug it in
Mehek14
  • Mehek14
yes
anonymous
  • anonymous
i got 98
Mehek14
  • Mehek14
\(h(t) = -16*1.41^2 + 45*1.41 + 40\\ 1.41^2=1.9881*-16=-31.81\\45*1.41=63.45\\63.45-31.81=31.64\\31.64+40=71.64\) 71.64 can be rounded to 72
anonymous
  • anonymous
ohh okay.
anonymous
  • anonymous
thank you!

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