Question

Jon is launching rockets in an open field. The path of the rocket can be modeled by the quadratic function h(t) = -16t^2 + 96t, where h(t) is the height in feet any time t in seconds. After how many seconds will the rocket reach a height of 80 feet for the second time

Answer 1

It will reach 80 feet for the second time after 5 seconds.

We set the equation equal to 80 feet:
80= -16t² + 96t

When solving quadratics, we want it equal to 0, so we subtract 80 from both sides:
80-80= -16t² + 96t - 80
0= -16t² + 96t - 80

We use the quadratic formula to solve:
$t=\frac{-96\pm \sqrt{96^2-4(-16)(-80)}}{2(-16)} \\ \\t=\frac{-96\pm \sqrt{9216-5120}}{-32} \\ \\t=\frac{-96\pm \sqrt{4096}}{-32} \\ \\t=\frac{-96\pm 64}{-32} \\ \\t=\frac{-96+64}{-32} \text{ or } 0=\frac{-96-64}{-32} \\ \\t=1 \text{ or } t=5$