Question

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 s(t) = -16t2 48t 64, where s is the height of the rocket and t is the time in seconds since the launch. After how many seconds will the rocket hit the ground?

Answer 1

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.