Posted by **Jane** on Saturday, February 2, 2013 at 5:21pm.

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.

- physics -
**Steve**, Saturday, February 2, 2013 at 5:27pm
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.

- physics -
**Damon**, Saturday, February 2, 2013 at 5:32pm
(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

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