Lecture12

Lecture12 - PHYSICS 149 Lecture 12 Chapter 4 4.4 Motion of...

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PHYSICS 149: Lecture 12 • Chapter 4 – 4.4 Motion of Projectiles – 4.5 Apparent Weight – 4.6 Air Resistance Lecture 12 Purdue University, Physics 149 1
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Midterm Exam 1 • Wednesday, February 23, 18:30 – 19:30 • Place: PHY 333 • Chapters 1 - 4 • The exam is closed book. • The exam is a multiple-choice test. • There will be ~15 multiple-choice problems. – maximum possible score will be 150 points. • Note that total possible score for the course is 1,000 points (see the course syllabus) • The difficulty level is about the same as the level of textbook problems. • You may make a single crib sheet – you may write on both sides of an 8.5” × 11.0” sheet Lecture 11 2 Purdue University, Physics 149
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Lecture 12 Purdue University, Physics 149 3 You and a friend are standing on level ground, each holding identical baseballs. At exactly the same time, and from the same height, you drop your baseball without throwing it while your friend throws her baseball horizontally as hard as she can. Which ball hits the ground first? A) Your ball B) Your friends ball C) They both hit the ground at the same time ILQ 1 When ball hits depends on y only! y(t) = y 0 + v yo + ½ a y t Same for both balls! v 0y = 0 and y = 0 therefore t=sqrt(2y 0 /g) result is independent of v 0x
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Lecture 12 Purdue University, Physics 149 4 ILQ 2 Fred throws a ball 30 mph vertically upward and then catches it again at the same height he threw it from. What is the speed of the ball when he catches it? (Neglect air resistance) A) v < 30 mph B) v = 30 mph C) v > 30 mph v y 2 = v y0 2 - 2g(y-y 0 ) v y 2 = v y0 2
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Motion with a Constant Acceleration Lecture 12 Purdue University, Physics 149 5 • All of the following examples are motions with a constant acceleration (if we ignore air resistance and if an object moves near the surface of Earth). – Free Fall – Motion of an Object Thrown Vertically Downward – Motion of an Object Thrown Vertically Upward – Motion of Projectiles a = g = 9.8 m/s 2 , downward F=mg, downward
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Lecture 12 Purdue University, Physics 149 6 y x y = y 0 + v y0 t - 1/2 gt 2 v y = v y0 -gt v y 2 = v y0 2 - 2g(y-y 0 ) Free Fall • Only force acting on object is GRAVITY –New ton s 2 nd Law Σ F y = ma y – Force is Weight = mg (near surface of earth) • -mg = ma y •a y = -g (- sign tells us it is in – y direction or down) • Acceleration is always g downwards – Velocity may be positive, zero or negative – Position may be positive, zero or negative
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Lecture 12 Purdue University, Physics 149 7 Free Fall • At the earth s surface freely falling bodies experience a constant acceleration of 9.8 m/s 2 y There is a choice of axes: 1) Choose positive y up in which case a=g=-9.8 m/s 2 2) Choose positive y down in which case a=g=9.8 m/s 2 g=9.8 m/s 2
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Motion in 2-D • X and Y are INDEPENDENT! • Break 2-D problem into two 1-D problems
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This note was uploaded on 04/23/2011 for the course PHYS 149 taught by Professor Staff during the Spring '08 term at Purdue.

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Lecture12 - PHYSICS 149 Lecture 12 Chapter 4 4.4 Motion of...

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