SOLUTION: I'm stumped about this problem. A rocket is fired into the air. The height of the rocket h(t) after t seconds is given by the formula h(t)=80t-16^2. After how many seconds is the

Algebra ->  Quadratic Equations and Parabolas  -> Quadratic Equations Lessons  -> Quadratic Equation Lesson -> SOLUTION: I'm stumped about this problem. A rocket is fired into the air. The height of the rocket h(t) after t seconds is given by the formula h(t)=80t-16^2. After how many seconds is the       Log On


   

Question 64225This question is from textbook Intermediate Algebra through applications
: I'm stumped about this problem. A rocket is fired into the air. The height of the rocket h(t) after t seconds is given by the formula h(t)=80t-16^2. After how many seconds is the rocket 96 feet high
This is what I started:
h(t)=80-16^2
h(96)=80(96)-16(96)^2 Now I am lost???? Can you show me how to complete this problem? Thank you so Much! This question is from textbook Intermediate Algebra through applications

Answer by 303795(602) About Me  (Show Source):
You can put this solution on YOUR website!
You have not written down the formula correctly
It is not (h(t)=80-16^2) but rather(h(t)=80t-16t^2)
You have also confused the time which is t with the height which is h(t) because the height varies with time.
You have to find the time (t)
h(t) means the height at that time t. You are told that it is 96.
so h%28t%29=80t-16t%5E2
96=80t-16t%5E2
0=80t-16t%5E2-96
0=-16t%5E2%2B80t-96
Multiply both sides by -1
0=16t%5E2-80t%2B96
Divide both sides by 16
0=t%5E2-5t%2B6
Factorise
0=%28t-2%29%28t-3%29
so t=2 or t=3
So the rocket is at 96 feet after 2 seconds and again at 96 feet after 3 seconds as it comes back down again.