problem02_91 solution

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

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2.91: a) The final speed of the first part of the fall (free fall) is the same as the initial speed of the second part of the fall (with the Rocketeer supplying the upward acceleration), and assuming the student is a rest both at the top of the tower and at the ground, the distances fallen during the first and second parts of the fall are g v g v 10 and 2 2 1 2 1 , where v 1 is the student's speed when the Rocketeer catches him. The distance fallen in free fall is then five times the distance from the ground when caught, and so the distance above the ground when caught is one-sixth of the height of the tower, or 92.2 m.
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Unformatted text preview: b) The student falls a distance 6 5 H in time , 3 5 g H t = and the Rocketeer falls the same distance in time t–t , where t =5.00 s (assigning three significant figures to t is more or less arbitrary). Then, ). ( 2 1 ) ( 6 5 or , ) ( 2 1 ) ( 6 5 2 t t g t t H v t t g t t v H---=-+-= At this point, there is no great advantage in expressing t in terms of H and g algebraically; s 698 4 s 00 . 5 s m 29.40 m) 553 ( 5 2 . t t =-=-, from which . s m 1 . 75 = v c)...
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