problem02_86 solution

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

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2.86: a) The helicopter accelerates from rest for 10.0 s at a constant 2 s m 5.0 . It thus reaches an upward velocity of s m 50.0 s) 0 . 10 )( s m 0 . 5 ( 2 0 = = + = t a v v y y y and a height of m 250 s) 0 . 10 )( s m 0 . 5 ( 2 2 2 1 2 2 1 = = = t a y y at the moment the engine is shut off. To find the helicopter's maximum height use ) ( 2 0 2 0 2 y y a v v y y y - + = Taking m 250 0 = y , where the engine shut off, and since 0 2 = y v at the maximum height: m 378 ) s m 8 . 9 ( 2 s) m 0 . 50 ( m 250 2 2 2 max 2 0 0 max = - - = - = - y g v y y y or 380 m to the given precision. b) The time for the helicopter to crash from the height of 250 m where Powers stepped out and the engine shut off can be found from: 0 ) s m 8 . 9 ( s) m 0 . 50 ( m 250 2 1 2 2 2 1 2 0 0 = - + + = + + = t t t a t v y y y y where we now take the ground as 0 = y . The quadratic formula gives solutions of s 67 . 3 = t and 13.88 s, of which the first is physically impossible in this situation. Powers'
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Unformatted text preview: position 7.0 seconds after the engine shutoff is given by: m 9 . 359 ) s . 49 )( s m 8 . 9 ( 2 1 s) . 7 )( s m . 50 ( m 250 2 2 =-+ + = y at which time his velocity is s m 6 . 18 ) s . 7 )( s m 80 . 9 ( s m . 50 2-=-+ = + = gt v v y y Powers thus has s 88 . 6 . 7 88 . 13 =-more time to fall before the helicopter crashes, at his constant downward acceleration of 2 s m 2.0 . His position at crash time is thus: 2 2 1 t a t v y y y y + + = 2 2 ) s 88 . 6 )( s m . 2 ( 2 1 s) 88 . 6 )( s m 6 . 18 ( m 9 . 359-+-+ = m 6 . 184 = or 180 m to the given precision....
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This document was uploaded on 02/04/2008.

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