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Question:

A brick is dropped from a high scaffold. What is its velocity after 2.9 s?

PLEASE HELP!!!

Answer:

v=at (assuming acceleration is constant as it is in this case) acceleration is gravity, G=9.8 m/s or 10 m/s v=(10 m/s)2.9 29 m/s
v = u + at u = 0; a = 9.8m/s2 t = 2.9s v = 0 + 9.8 * 2.9 = 28.42m/s
A brick is dropped from a high scaffold. What is its velocity after 2.9 s? As an object fall, neglecting air resistance, its velocity increases 9.8 m/s each second. Final velocity = Initial velocity + 9.8 * time Initial velocity = 0 m/s Final velocity = 0 + 9.8 * 2.9 Final velocity = 28.42 m/s

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