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.
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.
2 answers
(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
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