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CHAPTER 2 - 16

CHAPTER 2 - 16 - 75(a , childfrom v =v 2 2 02 2ah 2 2 2 2 2...

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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  up to the point where the front of the train passes

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