Lect02 - Physics 211: Lecture 2 Todays Agenda Recap of 1-D...

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Physics 211: Lecture 2, Pg 1 Physics 211: Lecture 2 Physics 211: Lecture 2 Today Today ’s Agenda s Agenda z Recap of 1-D motion with constant acceleration z 1-D free fall has constant acceleration… z Review of Vectors z Halftime z 3-D Kinematics ¾ Shoot the monkey ¾ Baseball ¾ Independence of x and y components
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Physics 211: Lecture 2, Pg 2 Review: Review: z For constant acceleration we found: at v v 0 + = 2 0 0 at 2 1 t v x x + + = const a = x a v t t t v) (v 2 1 v ) x 2a(x v v 0 av 0 2 0 2 + = = z From which we derived:
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Physics 211: Lecture 2, Pg 3 Recall what you saw: Recall what you saw: 1 2 3 4 2 0 0 at 2 1 t v x x + + =
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Physics 211: Lecture 2, Pg 4 1 1 - - D Free D Free - - Fall Fall z This is a nice example of constant acceleration (gravity): z In this case, acceleration is caused by the force of gravity: ¾ Usually pick y -axis “upward” ¾ Acceleration of gravity is “down” : y a y = g y0 y v= v - g t 2 y 0 0 t g 2 1 t v y y + = y a v t t t g a y =
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Physics 211: Lecture 2, Pg 5 Gravity facts: Gravity facts: z g does not depend on the nature of the material! ¾ Galileo (1564-1642) figured this out without fancy clocks & rulers! ¾ On the surface of the earth, gravity acts to give a constant acceleration z demo - feather & penny in vacuum z Nominally, g = 9.81 m/s 2 ¾ At the equator g = 9.78 m/s 2 ¾ At the North pole g = 9.83 m/s 2 z More on gravity in a few lectures! Penny & feather
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Physics 211: Lecture 2, Pg 6 Problem: Problem: z The pilot of a hovering helicopter drops a lead brick from a height of 1000 m . How long does it take to reach the ground and how fast is it moving when it gets there? (neglect air resistance) 1000 m
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Physics 211: Lecture 2, Pg 7 Problem: Problem: z First choose coordinate system. ¾ Origin and y -direction. z Next write down position equation: z Realize that v 0y = 0 . 2 0y 0 gt 2 1 t v y y + = 2 0 gt 2 1 y y = 1000 m y = 0 y = 2 0 1 2 yg t
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Physics 211: Lecture 2, Pg 8 Problem: Problem: z Solve for time t when y = 0 given that y 0 = 1000 m. z Recall: z Solve for v y : y 0 = 1000 m y s 3 14 s m 81 9 m 1000 2 g y 2 t 2 0 . . = × = = = 2 0 1 2 yg t y = 0 ) ( 0 2 y 0 2 y y y a 2 v v - - = s m 140 gy 2 v 0 y / = ± =
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Physics 211: Lecture 2, Pg 9 Lecture 2, Lecture 2, Act 1 Act 1 1D free fall 1D free fall z Alice and Bill are standing at the top of a cliff of height Alice and Bill are standing at the top of a cliff of height H . Both throw a ball with initial speed . Both throw a ball with initial speed v 0 , Alice straight , Alice straight down down and Bill straight and Bill straight up up . The speed of the balls when . The speed of the balls when they hit the ground are they hit the ground are v A and and v B respectively. respectively. Which Which of the following is true: of the following is true: (a) (a) v A < < v B (b) (b) v A = = v B (c) (c) v A > > v B v 0 v 0 Bill Bill Alice Alice H v A v B
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Physics 211: Lecture 2, Pg 10 Lecture 2, Lecture 2, Act 1 Act 1 1D Free fall 1D Free fall z Since the motion up and back down is symmetric, intuition should tell you that v = v 0 ¾ We can prove that your intuition is correct: v 0 Bill Bill H v = v = v 0 This looks just like Bill threw This looks just like Bill threw the ball down with speed the ball down with speed v 0 , so , so
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Lect02 - Physics 211: Lecture 2 Todays Agenda Recap of 1-D...

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