If we ignore air resistance, a falling body will fall 16t2 feet in t seconds. What is the average velocity between t
1 answer:
Answer:
<em>262.4 m/s</em>
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Explanation:
The complete question is
If we ignore air resistance, a falling body will fall 16t^2 feet in t seconds. What is the average velocity between t=8 and t=8.4? Round your answer to two decimal places if necessary.
The distance fallen s = 16t^2
The velocity v = = 32t
If we substitute the values of t into the velocity v, we'll have
at t = 8 s, V1 = 32 x 8 = 256 m/s
at t = 8.4 s, V2 = 32 x 8.4 = 268.8 m/s
Average velocity = (V2 - V1)/2 = (268.8 + 256)/2 = <em>262.4 m/s</em>
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Answer:
h = 3.3 m (Look at the explanation below, please)
Explanation:
This question has to do with kinetic and potential energy. At the beginning (time of launch), there is no potential energy- we assume it starts from the ground. There, is, however, kinetic energy
Kinetic energy = m
Plug in the numbers = (4.0)(
)
Solve = 2(64) = 128 J
Now, since we know that the mechanical energy of a system always remains constant in the absence of outside forces (there is no outside force here), we can deduce that the kinetic energy at the bottom is equal to the potential energy at the top. Look at the diagram I have attached.
Potential energy = mgh = (4.0)(9.8)(h) = 39.2(h)
Kinetic energy = Potential Energy
128 J = 39.2h
h = 3.26 m
h= 3.3 m (because of significant figures)
