CHAPTER 2 - 16 - 75.(a)

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D a y  = 0 v 0  = 0 L TRAIN 75.( a ) At the top of the motion the velocity is zero, so we find the maximum height of the second  child from v 2  =  v 02 2  + 2 ah 2 ;    0 = (5.0 m/s) 2  + 2(– 9.80 m/s 2 ) h 2   , which gives  h 2  = 1.28 m =       1.3 m . ( b ) If the first child reaches a height  h 1  = 1.5 h 2   , we find the initial speed from v 2  =  v 01 2  + 2 ah 1  =  v 01 2  + 2 a (1.5 h 2 ) =  v 01 2  – 1.5 v 02 2  = 0; v 01 2  = (1.5)(5.0 m/s) 2 , which gives  v 01  =        6.1 m/s . ( c ) We find the time for the first child from y  =  y 0  +  v 0 t  +  ! at 2   0 = 0 + (6.1 m/s) t  +  ! (– 9.80 m/s 2 ) t 2 which gives  t   = 0 (when the child starts up), and         t  = 1.2 s . 76. We use a coordinate system with the origin at the  initial position of the front of the train.  We can  find the acceleration of the train from the motion 
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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 University of California, Berkeley.

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CHAPTER 2 - 16 - 75.(a)

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