problem05_82

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

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5.82: We take the upward direction as positive. The explorer’s vertical acceleration is 2 s m 7 . 3 - for the first 20 s. Thus at the end of that time her vertical velocity will be s. m 74 s) 20 )( s m 7 . 3 ( 2 - = - = = at v y She will have fallen a distance m 740 s) 20 ( 2 s m 74 av - =  - = = t v d and will thus be m 460 740 1200 = - above the surface. Her vertical velocity must reach zero as she touches the ground; therefore, taking the ignition point of the PAPS as , 0 0 = y ) ( 2 0 2 0 2 y y a v v y - + = 2 2 2 0 2 0 2 s m 6.0 or s m 95 . 5 460 s) m 74 ( 0 ) ( 2 = - - - = - - = y y v v a y which is the vertical acceleration that must be provided by the PAPS. The time it takes to reach the ground is given by s 4 . 12 s m 95 . 5 ) s m 74 ( 0 2 0 = - - = -
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Unformatted text preview: = a v v t Using Newtons Second Law for the vertical direction N 1400 or N 5 . 1447 s m )) 7 . 3 ( kg)(5.95 150 ( ) ( 2 PAPSv PAPSv =--= + =-= = + g a m mg ma F ma mg F which is the vertical component of the PAPS force. The vehicle must also be brought to a stop horizontally in 12.4 seconds; the acceleration needed to do this is 2 2 s m 66 . 2 s 4 . 12 s m 33 =-=-= t v v a and the force needed is N 399 ) s m kg)(2.66 150 ( 2 PAPSh = = = ma F or 400 N, since there are no other horizontal forces....
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