Final answer:
The speed at the end of the upward acceleration phase is 60 m/s, the maximum height reached is 88.4 m, and the time after t=0 that the maximum height is reached is 6.1 s.
Step-by-step explanation:
(i) To find the speed at the end of the upward acceleration phase, we can use the equation:
vf = vi + at
Where:
vf = final velocity
vi = initial velocity
a = acceleration
t = time
Given that the initial velocity is 0 m/s, the acceleration is 30 m/s^2, and the time is 2.0 s, we can substitute the values into the equation:
vf = 0 + (30)(2.0) = 60 m/s
Therefore, the speed at the end of the upward acceleration phase is 60 m/s.
(ii) To find the maximum height reached, we can use the equation:
h = vi*t + (1/2)at^2
Where:
h = maximum height
vi = initial velocity
t = time
a = acceleration
Since the rocket is at rest at the maximum height, the final velocity is 0 m/s. Therefore, we can substitute the values into the equation:
0 = vi*(2.0) + (1/2)(-9.8)(2.0)^2
Solving for vi, we find:
vi = 88.4 m/s
The vertical component of the initial velocity is equal to the speed at the end of the upward acceleration phase. Therefore, the maximum height reached is 88.4 m.
(iii) To find the time after t=0 that the maximum height is reached, we can use the equation:
vf = vi + at
where vf, vi, a, and t are the same as in part (i).
Substituting the values into the equation, we have:
0 = 60 + (-9.8)t
Solving for t, we find:
t = 6.1 s
Therefore, the time after t=0 that the maximum height is reached is 6.1 s.