Question

a group of astronauts launched a model rocket from a platform. Its flight path is modeled by h= -4t^2+24t+13 where h is the height of the rocket above the ground in meters, and t is the time after the launch in seconds. How many seconds did the model rocket stay above the ground since it left the platform

Answer 1

Answer:

6.5 seconds

Step-by-step explanation:

Keep in mind that when $h=0$, this is the same height for both when the model rocket takes off and lands, so when the rocket lands, time is positive. Thus:

$h=-4t^2+24t+13\\\\0=-4t^2+24t+13\\\\t=\frac{-b\pm\sqrt{b^2-4ac}}{2a}\\ \\t=\frac{-24\pm\sqrt{24^2-4(-4)(13)}}{2(-4)}\\ \\t=\frac{-24\pm\sqrt{576+208}}{-8}\\\\t=\frac{-24\pm\sqrt{576+208}}{-8}\\\\t=\frac{-24\pm\sqrt{784}}{-8}\\\\t=\frac{-24\pm28}{-8}\\\\t=\frac{-24-28}{-8}\\ \\t=\frac{-52}{-8}\\ \\t=\frac{52}{8}\\\\t=6.5$

So, the amount of seconds that the model rocket stayed above the ground since it left the platform is 6.5 seconds

Answer 2

First find where height is = 0 because once it dips below 0 then it will be below ground.
So 0 = -4t^2+24t+13
Find roots:
0=(2t-13)(2t+1)
t= -0.5 or 6.5 and cause we can’t have negative seconds
T= 6.5