CHAPTER 2 - 13

# CHAPTER 2 - 13 - y =y 1 2(20.0m/s)t(9.80m/s)t...

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y 1  =  y 2 ; (20.0 m/s) t  +  ! (– 9.80 m/s 2 ) t 2  = (12.0 m/s)( t  – 1.00 s) +  ! (– 9.80 m/s 2 )( t  – 1.00 s) 2 which gives  t  = 9.38 s. We find the height from  y 1  =  y 0  +  v 01 t   ! at 2  = 0 + (20.0 m/s)(9.38 s) +  ! (– 9.80 m/s 2 )(9.38 s) 2  = – 244 m. This means they       never collide .      The rock, thrown later, returns to the ground before the ball  does.  To confirm this we find the time when the rock strikes the ground: y 2  =  y 0  +  v 02 ( t  – 1.00 s) +  ! a ( t  – 1.00 s) 2   0 = 0 + (12.0 m/s)( t  – 1.00 s) +  ! (– 9.80 m/s 2 )( t  – 1.00 s) 2 , which gives  t  = 3.45 s. At this time the position of the ball is y 1  =  y 0  +  v 01 t   ! at 2  = 0 + (20.0 m/s)(3.45 s) +  ! (– 9.80 m/s 2 )(3.45 s) 2  = 10.7 m. 65. We use a coordinate system with the origin at the ground, up positive, with  t 1  the time when the rocket reaches the  bottom of the window and  t 2  =  t 1  + 0.15 s the time when the rocket reaches the top of the window.  A very quick burn

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