High School

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A model rocket is launched vertically from a platform 64 feet above the ground. The height the rocket reaches during the flight is modeled by the equation [tex]s(t) = -16t^2 + 48t + 64[/tex], where [tex]s[/tex] is the height of the rocket and [tex]t[/tex] is the time in seconds since the launch.

After how many seconds will the rocket hit the ground?

Answer :

At ground level, the height is = 0. Therefore, s(t) = 0.
0 = -16t2 + 48t + 64
Divide both sides by -16.
0 = -16 (t2 - 3t - 4)
We are left with this equation: 0 = t2 - 3t - 4
Factor out.
0 = (t - 4) (t+1)
We solve each separately.
0 = (t-4)
t = 4
0 = (t+1)
t = -1
We are left with two answers but time cannot be negative so t = 4 seconds.

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