Answer:
At t=2
Step-by-step explanation:
Since h(t) represents the height of the rocket at time t, we can plug in the height of 96 feet and set that equal to the function h(x). This means we get:
[tex]96=-16t^2+80t[/tex]
This can now be solved for by doing the following:
[tex]96=-16t^2+80t\\0=-16t^2+80t-96\\[/tex]
From this point we now have quadratic to work with. Using the quadratic formula we can see that:
[tex]x=\frac{-80{\pm}\sqrt{80^2-4\left(-16\right)\left(-96\right)}}{2\left(-16\right)}=\frac{-80{\pm}\sqrt{256}}{-32}=\frac{-80{\pm}16}{-32}[/tex]
If we solve for both the plus version and minus version of the formula we get:
[tex]t=\frac{-80+16}{-32}= \frac{-64}{-32}=2\\ \\t=\frac{-80-16}{-32}= \frac{-96}{-32}=3[/tex]
So the rocket will be at a height of 96 feet after 2 and 3 seconds. However the question asks when it will reach the height so we choose the smaller time. Therefore the answer is t=2 or after 2 seconds
