CHAPTER 2 - 14

# CHAPTER 2 - 14 - 68(a (b (c 5/2 a=(2.0m/s(5.0s =4.5m/s...

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H + a 1 a 2 y 01  = 0 y 02  = 0 v 01  = 0 v 02  =  v 1 68. ( a ) We find the speed by integration: ( b ) We find the displacement by integration: ( c ) For the given time we have a  = (2.0 m/s 5/2 )(5.0 s) 1/2  =       4.5 m/s 2 ; v  = (10 m/s) +  % (2.0 m/s 5/2 )(5.0 s) 3/2  =       25 m/s ; x  = (10 m/s)(5.0 s) + (4/15)(2.0 m/s 5/2 )(5.0 s) 5/2  =       80 m . 69. The height reached is determined by the initial velocity.  We assume the same initial velocity of the object on the  Moon and Earth.  With a vertical velocity of 0 at the highest point, we have  v 2  =  v 0 2  + 2 ah ; 0 =   v 0 2  + 2(–  g ) h , so   we get  v 0 2  = 2 g Earth h Earth  = 2 g Moon h Moon   ,  or   h Moon  = ( g Earth / g Moon ) h Earth  =          6 h Earth . Page *

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70. For the falling motion, we use a coordinate system with the origin  at the fourth-story window and down positive.  For the stopping
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## This note was uploaded on 03/08/2010 for the course PHYSICS 7A/7B taught by Professor All during the Fall '08 term at Berkeley.

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CHAPTER 2 - 14 - 68(a (b (c 5/2 a=(2.0m/s(5.0s =4.5m/s...

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