for a freely falling object dropped from rest, what is the acceleration at the end of the fifth second of fall? the sixth second? at the end of any elapsed time?

I recall Galelio found it was constant, 9.8m/s^2 at any time.

To determine the acceleration of a freely falling object dropped from rest at the end of any given time interval, you need to consider the acceleration due to gravity, which is approximately 9.8 meters per second squared (m/s²) near the Earth's surface.

Let's break down the question into three parts:

1. The acceleration at the end of the fifth second of fall:
At the end of the fifth second, the object has been falling for 5 seconds. Since the acceleration due to gravity remains constant throughout the fall, the acceleration at the end of the fifth second is still -9.8 m/s² (negative, indicating it acts downward). This negative value shows that the object is accelerating downwards.

2. The acceleration at the end of the sixth second of fall:
At the end of the sixth second, the object has been falling for 6 seconds. Again, the acceleration due to gravity remains constant, so the acceleration at the end of the sixth second is -9.8 m/s². The object continues to accelerate downward.

3. The acceleration at the end of any elapsed time:
If you want to find the acceleration at the end of any other elapsed time, the value will still be -9.8 m/s². This is because, in free fall, the acceleration is constant and solely determined by gravity on Earth.

Remember, the acceleration is the rate of change of velocity, so in free fall, the velocity of the object keeps increasing at a constant rate of -9.8 m/s².