This document discusses the physics of falling objects. It defines physics as the study of matter, energy, and their interactions. It states that near the Earth's surface, in the absence of friction, all objects experience a constant downward acceleration due to gravity of about 9.81 m/s2. It provides the kinematic equations for analyzing vertical motion under constant acceleration due to gravity and emphasizes the importance of sign conventions in such problems. Finally, it works through an example problem of calculating the distance fallen by a ball dropped from a tower over different time intervals.

1. Falling Objects

2. What is physics exactly? The branch of science concerned with the nature and properties of matter and energy.The study of the physical worldIncludes mechanics, heat, light and other radiation, sound, electricity, magnetism, and the structure of atoms

3. All objects “fall” at same rate near the surface of the earth, & in the absence of frictionObjects will fall at constant acceleration (at the rate of acceleration due to gravity)By “fall”, we are referring to the rate at which velocity changesAnd I’m free…..free falling!

4. SI units: ay = g = -9.81 m/s2The velocity of an object changes 9.81 m/s every second. English units: ay = g = -32.2 ft/s2The velocity of an object changes 32.2 m/s every second. Acceleration due to gravity

5. Acceleration due to gravity

6. It NEVER ceases to existIt ALWAYS works DOWNIt is NEVER zeroAcceleration due to gravity

7. For vertical motion, we use the same kinematic equations as before: In your book: xi, xf, Δx become yi, yf, ΔyIn your book: Acceleration (a) is gAnalyzing falling objects using kinematic equations

8. Negative & positive signs are VERY important in vertical motionMUST be consistentG is negativeVertical downward displacement away from origin is negativeVertical upward displacement away from origin is positiveAcceleration due to gravity

9. A ball is dropped from a tower 70.0 m high. If we neglect air friction, determine how far the ball has fallen after:1.00 s2.00 s3.00 sProblem Solving Strategy:Identify knowns:Xi  yiXi  YfViVfa  g tChoose an equationPlug in values to solve the problem Example: xf = xi + vit + 0.5at2  yf= yi+ vit + 0.5gt2yf =0 + 0+ 0.5(-9.81)t2yf at 1.00 s = -4.90 myf at 2.00 s = -19.6 myf at 3.00 s = -44.1 m