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Mathematics 18 Online
OpenStudy (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.

OpenStudy (anonymous):

then After 3 seconds, the rocket is traveling at a velocity of _________ feet per second.

OpenStudy (lgbasallote):

Can I just answer your first question? Because it is..lol xD

OpenStudy (anonymous):

lol sure!!

OpenStudy (anonymous):

s(3)=-16(3)^2+80(3)=96 ft

OpenStudy (anonymous):

v(t)=s'(t)=-32t+80 v(3)=-32(3)+80=-16ft/s

OpenStudy (anonymous):

thanks

OpenStudy (anonymous):

wait how do I find if The rocket is 60 feet above ground at what times?

OpenStudy (anonymous):

set s(t)=60,solve for t

OpenStudy (anonymous):

you'll have a quadratic equation, it's a parabola so there should be 2 answers

OpenStudy (anonymous):

ok i got -38,200

OpenStudy (anonymous):

and the the other answer is pos. 38,200

OpenStudy (anonymous):

am I right?

OpenStudy (anonymous):

am I right?

OpenStudy (anonymous):

let me check

OpenStudy (anonymous):

i highly doubt those are the right answers, the rocket will have hit the ground long before those times

OpenStudy (anonymous):

\[-16t^2+80x-60=0\] Solving for t, \[t= \frac{(5 \pm \sqrt{10})}{2}\]

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