CHAPTER 3 - 10

# CHAPTER 3 - 10 - 26 , verticalmotion y=y v t!a t 0 0y y 2...

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x θ y v 0 O R 26. 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 tiger to reach the ground from its  vertical motion: y  =  y 0  +  v 0 y t  +  ! a y t 2 ;   6.5 m = 0 + 0   ! (9.80 m/s 2 ) t 2 , which gives  t  = 1.15 s. The horizontal motion will have constant velocity.   We find the distance from the base of the rock from x  =  x 0  +  v 0 x t x  = 0 + (4.0 m/s)(1.15 s) =          4.6 m . 27. 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 height of the cliff from the vertical displacement:  y  =  y 0  +  v 0 y t  +  ! a y t 2 ;   y  = 0 + 0   ! (9.80 m/s 2 )(3.0 s) 2  =         44 m .

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CHAPTER 3 - 10 - 26 , verticalmotion y=y v t!a t 0 0y y 2...

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