ch04-p049 - 49. (a) The skier jumps up at an angle of 0 =...

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which gives sin (7.27 m)sin(11.3 ) 1.42 m. yd α =− ° =− Therefore, at landing the skier is approximately 1.4 m below the launch level. (c) The time it takes for the skier to land is 00 cos (7.27 m)cos(11.3 ) 0.72 s cos (10 m/s)cos(9.0 ) x xd t vv θ ° == = = ° . Using Eq. 4-23, the x -and y -components of the velocity at landing are 2 cos (10 m/s)cos(9.0 ) 9.9 m/s sin (10m/s)sin(9.0 ) (9.8m/s )(0.72s) 5.5m/s x y g t ° = = ° = Thus, the direction of travel at landing is 11 tan 29.1 . 9.9m/s y x v v −− §· = ° ¨¸ ©¹ or 29.1 ° below the horizontal. The result implies that the angle between the skier’s path and the slope is 29.1 11.3 17.8 φ ° , or approximately 18 ° to two significant figures. 49. (a) The skier jumps up at an angle of 0 9.0 up from the horizontal and thus returns to the launch level with his velocity vector 9.0 ° below the horizontal. With the snow surface making an angle of
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