## 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.

1. anonymous

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

2. anonymous

3. anonymous

lol sure!!

4. anonymous

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

5. anonymous

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

6. anonymous

thanks

7. anonymous

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

8. anonymous

set s(t)=60,solve for t

9. anonymous

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

10. anonymous

ok i got -38,200

11. anonymous

and the the other answer is pos. 38,200

12. anonymous

am I right?

13. anonymous

am I right?

14. anonymous

let me check

15. anonymous

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

16. anonymous

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