CHAPTER 3 - 24 - 12m=1.0m+v sin40t+! (9.80m/s )t . 0

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x h y v 0 O L 12 m = 1.0 m +  v 0   sin 40°  t  +  ! (– 9.80 m/s 2 ) t 2 . We can use the first equation to eliminate  v 0 t   from the second and  solve for  t , which gives  t  = 3.67 s.   When this value is used in the first equation, we get  v 0  =        33 m/s . 83. We choose a coordinate system with the origin at the takeoff point, with  x  horizontal and  y  vertical, with the  positive direction down.   We find the time for the diver to reach the water from the vertical motion: y  =  y 0  +  v 0 y t  +  ! a y t 2 ;   35 m = 0 + 0   ! (9.80 m/s 2 ) t 2 , which gives        t  = 2.7 s . The horizontal motion will have constant velocity.   We find the minimum horizontal initial velocity needed to land beyond the rocky outcrop from x  =  x 0  +  v 0 x t 5.0 m = 0 +  v 0 (2.7 s), which gives       v 0  = 1.9 m/s .
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CHAPTER 3 - 24 - 12m=1.0m+v sin40t+! (9.80m/s )t . 0

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