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# Lect02 - Physics 211 Lecture 2 Todays Agenda q q q q Recap...

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Physics 211: Lecture 2, Pg 1 Physics 211: Lecture 2 Physics 211: Lecture 2 Today’s Agenda Today’s Agenda Recap of 1-D motion with constant acceleration 1-D free fall has constant acceleration… Review of Vectors Halftime 3-D Kinematics Shoot the monkey Baseball Independence of x and y components

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Physics 211: Lecture 2, Pg 2 Review: Review: 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 + = - = - From which we derived:
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-D Free-Fall 1-D Free-Fall This is a nice example of constant acceleration (gravity): In this case, acceleration is caused by the force of gravity: Usually pick y -axis “upward” Acceleration of gravity is “down” : y a y = - g y 0y v = v - gt 2 y 0 0 t g 2 1 t v y y - + = y a v t t t g a y - =
Physics 211: Lecture 2, Pg 5 Gravity facts: Gravity facts: g does not depend on the nature of the material! Galileo (1564-1642) figured this out without fancy clocks & rulers! On the surface of the earth, gravity acts to give a constant acceleration Nominally, g = 9.81 m/s 2 At the equator g = 9.78 m/s 2 At the North pole g = 9.83 m/s 2 More on gravity in a few lectures! Penny & feather

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Physics 211: Lecture 2, Pg 6 Problem: Problem: 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
Physics 211: Lecture 2, Pg 7 Problem: First choose coordinate system. Origin and y -direction. Next write down position equation: 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 y gt

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Physics 211: Lecture 2, Pg 8 Problem: Problem: Solve for time t when y = 0 given that y 0 = 1000 m. Recall: 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 y gt y = 0 ) ( 0 2 y 0 2 y y y a 2 v v - - = s m 140 gy 2 v 0 y / - = ± =
Lecture 2, Lecture 2, Act 1 Act 1 1D free fall 1D free fall 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 v0 , 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 vA and and vB respectively.

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Lect02 - Physics 211 Lecture 2 Todays Agenda q q q q Recap...

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