178k views
0 votes
A model rocket is launched straight up into the air. The height y (in feet) of the rocket after t seconds can be modeled by y=-16t^2+128t. How many seconds is the rocket in the air?

User Jalpesh
by
7.7k points

1 Answer

1 vote

Answer:

8

Explanation:

To find the time the rocket is in the air, we can find the time when the height is 0 (remember this time is not 0).


-16t^2+128t=0 \\ \\ t^2-8t=0 \\ \\ t(t-8)=0 \\ \\ t=0,8 \\ \\ t \\eq 0 \implies t=8

User Palak Arora
by
7.0k points