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Answer :
Final answer:
The rocket's original acceleration during the first 16 seconds of the flight is 395.75 m/s².
Explanation:
The rockets original acceleration during the first 16 seconds of the flight can be determined by breaking down the motion into two phases: the phase when the rocket is accelerating and the phase when the rocket is freely moving. In the first phase, the rocket's acceleration can be calculated using the kinematic equation: v = u + at.
Since the rocket starts from rest, the initial velocity u is 0 m/s. The final velocity v at the end of 12 seconds is given by
v = at.
The final velocity is also the velocity of the rocket at the start of the second phase. In the second phase, the rocket is moving freely under the influence of gravity, so its velocity decreases at a rate of g = 9.8 m/s².
Therefore, the rockets original acceleration during the first 16 seconds of the flight is equal to the acceleration during the first phase, which can be calculated as a = v / t
Given that the altitude of the rocket at t = 21 seconds is 4749 meters, we can calculate the final velocity v at the end of the first phase using the equation v = at.
Substituting the values, we get 4749 = a * 12. Rearranging the equation, we find a = 4749 / 12
= 395.75 m/s².
Therefore, the rockets original acceleration during the first 16 seconds of the flight is 395.75 m/s².
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