CHAPTER 2 - 10 - y=y +v t+!at ; 0 0 2...

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y  = 0 H a + v 0 y 0 y  =  y 0  +  v 0 t  +  ! at 2 0 = 0 + (7.07 m/s) t  +  ! (– 9.80 m/s 2 ) t 2 which gives  t   = 0 (when the kangaroo jumps), and         t  = 1.44 s . 52. We use a coordinate system with the origin at the ground and up positive. When the ball returns to the ground, its displacement is zero, so we have y  =  y 0  +  v 0 t  +  ! at 2 0 = 0 +  v 0 (3.1 s) +  ! (– 9.80 m/s 2 )(3.1 s) 2 , which gives        v 0  = 15 m/s . At the top of the motion the velocity is zero, so we find the height  h  from v 2  =  v 0 2  + 2 ah ;    0 = (15 m/s) 2  + 2(– 9.80 m/s 2 ) h , which gives         h  = 12 m . 53. We use a coordinate system with the origin at the ground and up positive.  We assume you can throw the object 4  stories high, which is about 12 m. We can find the initial speed from the maximum height (where the velocity is zero):
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CHAPTER 2 - 10 - y=y +v t+!at ; 0 0 2...

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