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anonymous
 4 years ago
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.
anonymous
 4 years ago
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.

This Question is Closed

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0then After 3 seconds, the rocket is traveling at a velocity of _________ feet per second.

lgbasallote
 4 years ago
Best ResponseYou've already chosen the best response.0Can I just answer your first question? Because it is..lol xD

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0s(3)=16(3)^2+80(3)=96 ft

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0v(t)=s'(t)=32t+80 v(3)=32(3)+80=16ft/s

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0wait how do I find if The rocket is 60 feet above ground at what times?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0set s(t)=60,solve for t

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0you'll have a quadratic equation, it's a parabola so there should be 2 answers

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0and the the other answer is pos. 38,200

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0i highly doubt those are the right answers, the rocket will have hit the ground long before those times

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0\[16t^2+80x60=0\] Solving for t, \[t= \frac{(5 \pm \sqrt{10})}{2}\]
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