Why acceleration is not 9.8m/s^2?

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SUMMARY

The average acceleration of an object dropped from a height of 39.4m, which hits the ground at a speed of 1.61m/s, is calculated to be 3.29×10-2 m/s2. This value differs from the standard gravitational acceleration of -9.8 m/s2 due to the choice of coordinate system and potential external forces acting on the object. The discussion highlights that the sign of acceleration is arbitrary and contingent upon the defined coordinate system.

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Homework Statement



An object is dropped from ret from a height of 39.4m. What is its average acceleration (assumed to be constant) if it hits the ground with a speed of 1.61m/s?


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The Attempt at a Solution



I actually got the answer from this question. But I am wondering why a is not -9.8m/s^2
The answer is 3.29×10-2 m/s^2, shouldn't it be negative as acceleration is acting downwards?
 
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Possible reasons:
- The object is not dropped on earth, but somewhere else
- Some other force acts on it

The sign of the acceleration is arbitrary and depends on the coordinate system.
 

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