Answer:
90 m
Explanation:
The rocket undergoes constant acceleration. We can use kinematic equations, also known as SUVAT equations, to model the rocket's motion. For this problem, we will use the following equations:
s = ut + ½ at²
v = u + at
v² = u² + 2as
where
During the first part of the rocket's motion, it starts from rest (u = 0) and accelerates upward at a = 10.0 m/s² for a time of t = 3.0 s. The displacement after this time is:
s = ut + ½ at²
s = (0) (3.0) + ½ (10.0) (3.0)²
s = 45 m
The velocity of the rocket after this time is:
v = u + at
v = 0 + (10.0) (3.0)
v = 30.0 m/s
Next, the rocket decelerates from this speed (u = 30.0 m/s) at a rate of a = -10.0 m/s² until it reaches a stop (v = 0 m/s). The displacement of the rocket at this point is:
v² = u² + 2as
(0)² = (30.0)² + 2 (-10.0) s
s = 45 m
The rocket travels up 45 meters during acceleration, then another 45 meters during deceleration. The total height reached is:
45 m + 45 m = 90 m
