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Answer:
About 1.85 seconds and 13.15 seconds.
Step-by-step explanation:
The height (in feet) of the rocket t seconds after launch is given by the equation:
[tex]h = -16t^2 + 240 t[/tex]
And we want to determine how many seconds after launch will be rocket be 390 feet above the ground.
Thus, let h = 390 and solve for t:
[tex]390 = -16t^2 +240t[/tex]
Isolate:
[tex]-16t^2 + 240 t - 390 = 0[/tex]
Simplify:
[tex]8t^2 - 120t + 195 = 0[/tex]
We can use the quadratic formula:
[tex]\displaystyle x = \frac{-b\pm\sqrt{b^2 -4ac}}{2a}[/tex]
In this case, a = 8, b = -120, and c = 195. Hence:
[tex]\displaystyle t = \frac{-(-120)\pm \sqrt{(-120)^2 - 4(8)(195)}}{2(8)}[/tex]
Evaluate:
[tex]\displaystyle t = \frac{120\pm\sqrt{8160}}{16}[/tex]
Simplify:
[tex]\displaystyle t = \frac{120\pm4\sqrt{510}}{16} = \frac{30\pm\sqrt{510}}{4}[/tex]
Thus, our two solutions are:
[tex]\displaystyle t = \frac{30+ \sqrt{510}}{4} \approx 13.15 \text{ or } t = \frac{30-\sqrt{510}}{4} \approx 1.85[/tex]
Hence, the rocket will be 390 feet above the ground after about 1.85 seconds and again after about 13.15 seconds.