Solved: In Exercises 7780, an object moving vertically is at the given heights at the
Chapter 8, Problem 79(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 = 352 feet.
At t = 2 seconds, s = 272 feet.
At t = 3 seconds, s = 160 feet.
Text Transcription:
s=frac{1}{2}at^2+v_0t+s_0
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