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An object is dropped from a height of 26.8 m. What is its average acceleration...
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The object is dropped from a height of 26.8 m. As it is dropped, the initial velocity of the object is 0 m/s. It is assumed that the acceleration of the object is a constant over the entire distance that it falls. It hits the ground with a speed 3.96 m/s.
The relation between initial velocity (u), final velocity (v), displacement of an object (s) and its acceleration a is:
v^2 - u^2 = 2*a*s
Substituting the values in the problem:
3.96^2 - 0 = 2*26.8*a
=> a = 0.2925 m/s^2
The average acceleration of the object is 0.2925 m/s^2
Posted by justaguide on February 13, 2013 at 5:40 PM (Answer #1)
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