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problem03_92

# University Physics with Modern Physics with Mastering Physics (11th Edition)

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3.92: The x -position of the plane is t ) m/s 236 ( and the x -position of the rocket is . ) ( 30 cos ) m/s 80 . 9 )( 00 . 3 ( 2 / 1 ) m/s 236 ( 2 2 T t t - ° + The graphs of these two have the form, If we take 0 = y to be the altitude of the airliner, then 2 2 2 ) )( 30 (sin ) m/s 80 . 9 )( 00 . 3 ( 2 / 1 ) ( 2 / 1 ) ( T t T t gT gT t y - ° + - - - = for the rocket. This graph looks like By setting 0 = y for the rocket, we can solve for t in terms of . ) )( m/s 35 . 7 ( ) ( ) m/s 80 . 9 ( ) m/s 90 . 4 ( 0 , 2 2 2 2 2 T t T t T T T - + - - - = Using the quadratic formula for the variable
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Unformatted text preview: T, t x-= we find ) m/s 35 . 7 ( 2 ) 9 . 4 )( m/s 35 . 7 )( 4 ( ) m/s 80 . 9 ( ) m/s 80 . 9 ( 2 2 2 2 2 2 T T T T t x + + =-= or . 72 . 2 T t = Now, using the condition that , x x m 1000 plane rocket =-we find m, 1000 ) m/s 236 ( ) ( m/s 7 . 12 ( m/s) 236 ( 2 2 =--× + t T t ) t or . s 6 . 78 ) 72 . 1 ( 2 2 = T Therefore s. 15 . 5 = T...
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