anonymous
  • anonymous
why is this question so difficult? The height of a rocket launched from the ground is given by the function s(t) = -16 t^2 + 80 t, where s is in feet above ground and t is in seconds. After 3 seconds, the rocket is ________ feet above ground.
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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chestercat
  • chestercat
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anonymous
  • anonymous
then After 3 seconds, the rocket is traveling at a velocity of _________ feet per second.
lgbasallote
  • lgbasallote
Can I just answer your first question? Because it is..lol xD
anonymous
  • anonymous
lol sure!!

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More answers

anonymous
  • anonymous
s(3)=-16(3)^2+80(3)=96 ft
anonymous
  • anonymous
v(t)=s'(t)=-32t+80 v(3)=-32(3)+80=-16ft/s
anonymous
  • anonymous
thanks
anonymous
  • anonymous
wait how do I find if The rocket is 60 feet above ground at what times?
anonymous
  • anonymous
set s(t)=60,solve for t
anonymous
  • anonymous
you'll have a quadratic equation, it's a parabola so there should be 2 answers
anonymous
  • anonymous
ok i got -38,200
anonymous
  • anonymous
and the the other answer is pos. 38,200
anonymous
  • anonymous
am I right?
anonymous
  • anonymous
am I right?
anonymous
  • anonymous
let me check
anonymous
  • anonymous
i highly doubt those are the right answers, the rocket will have hit the ground long before those times
anonymous
  • anonymous
\[-16t^2+80x-60=0\] Solving for t, \[t= \frac{(5 \pm \sqrt{10})}{2}\]

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