a rocket is launched in the air. Its height in feet is given by h(t)=-16t^2 +88t where t represents the time in seconds after the launch. how long is the rocket in the air?

Answer 1

The rocket is on the ground when the height equals zero so we need to solve:

0 = -16t² + 88t

0 = -8t(2t - 11)

When a•b = 0, the only way that can happen is if either a = 0 or if b = 0 so we take each part of our factored quadratic and set it equal to zero and solve"

-8t = 0 occurs when t = 0

2t - 11 = 0 occurs when 2 = 5.5

So the rocket is at ground level when t = 0 seconds and it is at ground level when t = 5.5 seconds and in between it is in the air somewhere. So the total time that it is in the air is 5.5 seconds.