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CHAPTER 3 - 12

# CHAPTER 3 - 12 - y=y v t!a t 0 0y 2 y down 2 down 0=h 0(g)t...

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We find the time to fall from the highest point from y  =  y 0  +  v 0 y t  +  ! a y t down 2 ;   0 =  h  + 0 +  ! (–  g ) t down 2  , which gives  t down  = (2 h / g ) 1/2   =  [2( v 0 y 2 /2 g )/ g ] 1/2   =  v 0 y / g , which is the same as  t up . 35. To plot the trajectory, we need a  relationship between  x  and  y , which can be obtained by eliminating  t  from the equations for the two components  of the motion: x  =  v 0 x t  =  v 0  (cos  θ ) t ; y  =  y 0  +  v 0 y t  +  ! a y t 2       = 0 +  v 0  (sin  θ ) t  +  ! (–  g ) t 2 . The relationship is y  = (tan  θ ) x  –  ! g ( x / v 0  cos  θ ) 2 . 36. The arrow will hit the apple at the same elevation, so we can use the expression for the horizontal range:  R  =  v 0 2   sin(2 θ 0 )/ g ; 25.0 m = (22.5 m/s) 2  sin(2 θ 0 )/(9.80 m/s 2 ), which gives  sin(2 θ 0 ) = 0.484,   or   2 θ 0  = 28.9 °  ,           θ 0  = 14.5 ° .

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CHAPTER 3 - 12 - y=y v t!a t 0 0y 2 y down 2 down 0=h 0(g)t...

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