Answer: In Exercises 7780, an object moving vertically is at the given heights at the
Chapter 8, Problem 80(choose chapter or problem)
Vertical Motion In Exercises 77–80, an object moving vertically is at the given heights at the specified times. Find the position equation \(s=\frac{1}{2}at^2+v_0t+s_0\) for the object.
At t = 1 second, s = 132 feet.
At t = 2 seconds, s = 100 feet.
At t = 3 seconds, s = 36 feet.
Text Transcription:
s=frac{1}{2}at^2+v_0t+s_0
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