Final answer:
The rocket's height after 10.0 seconds is calculated by integrating the given acceleration function to find velocity and then height, resulting in the rocket being at a height of 466.68 meters above the Earth's surface.
Step-by-step explanation:
To calculate the height of a rocket after 10.0 seconds given the vertical acceleration equation ay = (2.80 m/s³)t, we must integrate the acceleration function twice to find the velocity function and then the height function. Starting from rest means that initial velocity voy is zero.
Firstly, we integrate the acceleration to find velocity:
- ∫ ay dt = ∫ (2.80 m/s³)t dt = (2.80 m/s³)(t²/2) + C
Since the rocket starts from rest, the constant C is zero, making the velocity function:
vy(t) = (2.80 m/s³)(t²/2)
Next, we integrate the velocity to find the height:
- ∫ vy(t) dt = ∫ (2.80 m/s³)(t²/2) dt = (2.80 m/s³)(t³/6) + C'
In this case, the constant C' represents the initial height, which is zero since the rocket starts from the surface of the Earth, leaving us with the height function:
y(t) = (2.80 m/s³)(t³/6)
Substituting t = 10.0 s into our height function:
y(10.0 s) = (2.80 m/s³)((10.0 s)³/6) = (2.80 m/s³)(1000 s³/6) = (2.80 m/s³)(166.67 s³) = 466.68 meters
Therefore, the rocket reaches a height of 466.68 meters above the Earth's surface after 10.0 seconds.