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?

Answers

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
To solve the problem, substitute 80 feet to value of h(t). so given that
h(t) = -16t^2 + 96t

so
h(t) = -16t^2 + 96t = 80
 -16t^2 + 96t - 80 = 0
divide the whole equation with -16
(  -16t^2 + 96t - 80 = 0) 1/-16
t^2 - 6t + 5 = 0
factor the quadratic equation
( t - 5 ) ( t - 1 ) = 0
t = 5
t = 1
so the answer would be 5 s, because it the rocket will reach the 80 ft for the first time at 1 s


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