CHAPTER 3 - 16 - 50....

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50. We find the time of flight from the vertical motion: y  =  y 0  +  v 0 y t  +  ! a y t 2 ;   0 = 0 + ( v 0   sin  θ 0 ) t  +  ! (–  g ) t 2 which gives  t  = 2 v 0   sin  0 / g  = 2(32 m/s) sin 55°/(9.80 m/s 2 ) = 5.35 s. The horizontal distance the ball travels is x   x 0  +  v 0 x t  = 0 + ( v 0   cos  ) t   = ( v 0   cos  ) t   = (32 m/s) cos 55° (5.35 s) = 98.2 m. From the top view of the positions as the outfielder runs  from  A  to  B , we see that d 1  =  x  sin  φ  = (98.2 m) sin 22° = 36.8 m; d 2  =  L  –  x  cos   = 85 m – (98.2 m) cos 22° = – 6.0 m. Thus the angle from the line to the plate that the  outfielder must run is given by tan  α  =  d 1 / d 2  = (36.8 m)/(– 6.0 m) = – 6.13,   = 99.3°. The distance is  d  = ( d 1 2  +  d 2 2 ) 1/2  = [(36.8 m) 2  + (– 6.0 m) 2 ] 1/2  = 37.3 m, so his speed is
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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 3 - 16 - 50....

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