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?

My quesion is, is a not 9.8 since gravity is acting on the object.

v = √(2as)

1.61 = √(2*a*39.4)
a 0.0329

Obviously not on earth, or there is a major updraft.

On earth the velocity would be much more than 1.61 m/s.

(1/2) m v^2 = m g h

so
v^2/2 = 1.61^2 / 2 = 1.296 m^2/s^2

so
g = 1.296/39.4
= .0329 m/s^2
must be a balloon :)

alternate way
if acceleration is constant then average v = 1.61/2 = .805 m/s
time = 39.4 / .805 = 48.9 seconds
change in v/time = average acceleration = 1.61 / 48.9 = .0329 m/s^2 same answer

Yes, you are correct. The acceleration due to gravity, denoted as "g," is approximately equal to 9.8 m/s² on the surface of the Earth. However, in this question, we need to calculate the average acceleration of the object as it falls.

To solve this problem, we can use the kinematic equation:

v² = u² + 2as

where:
- v is the final velocity (1.61 m/s in this case)
- u is the initial velocity (0 m/s since the object was dropped from rest)
- a is the average acceleration (what we need to find)
- s is the displacement or height (39.4 m in this case, as the object was dropped from that height)

Rearranging the equation to solve for a, we get:

a = (v² - u²) / (2s)

Plugging in the values:

a = (1.61 m/s)² - (0 m/s)² / (2 * 39.4 m)

Simplifying, we find:

a = 2.59521 / 78.8

a ≈ 0.0329 m/s²

Therefore, the average acceleration of the object as it falls is approximately 0.0329 m/s².