Question

A rocket was launched, and its height, ℎ, in meters, above the ground after time, t, in seconds, is represented by h=11+10t−2t^2 . For how many seconds was the rocket in the air?

A. 3 seconds
B. 4 seconds
C. 5 seconds
D. 6 seconds

Answer 1

The rocket was in the air for 3.79 seconds.

Step-by-step explanation:

To determine how long the rocket was in the air, we need to find the time at which the rocket's height reaches zero. We can use the given equation h=11+10t-2t^2, where h is the height in meters and t is the time in seconds.

Setting the equation equal to zero, we get 11+10t-2t^2=0. By solving this quadratic equation, we find two possible solutions: t=3.79 seconds and t=0.54 seconds.

Since the rocket's height reaches zero twice during its trajectory, we take the longer solution, t=3.79 seconds, to determine the time the rocket was in the air.