Answer:
To find the total time that the rocket is in the air, we can split the motion into two parts: the time it takes to reach the altitude of 415 m while accelerating, and the time it takes to fall back to the ground after the engines are turned off.
First, let's find the time it takes to reach the altitude of 415 m while accelerating.
We can use the kinematic equation for motion with constant acceleration:
\[ d = v_i t + \frac{1}{2} a t^2 \]
Where:
- \( d \) is the displacement (415 m),
- \( v_i \) is the initial velocity (0 m/s),
- \( a \) is the acceleration (20.0 m/s²),
- \( t \) is the time.
Plugging in the values, we get:
\[ 415 = 0 \times t + \frac{1}{2} \times 20.0 \times t^2 \]
Solving for \( t \), we get:
\[ 415 = 10 t^2 \]
\[ t^2 = \frac{415}{10} \]
\[ t^2 = 41.5 \]
\[ t = \sqrt{41.5} \]
\[ t \approx 6.44 \text{ seconds} \]
So, it takes approximately 6.44 seconds for the rocket to reach the altitude of 415 m while accelerating.
Now, to find the total time the rocket is in the air, we need to double this time since the rocket will spend the same amount of time coasting downward under gravity after the engines are turned off.
Therefore, the total time the rocket is in the air is approximately \( 2 \times 6.44 \) seconds, which is approximately \( 12.88 \) seconds.
