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40_Physics ProblemsTechnical Physics

40_Physics ProblemsTechnical Physics - Chapter 2 P2.50 Take...

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Chapter 2 41 P2.50 Take downward as the positive y direction. (a) While the woman was in free fall, y = 144 ft , v i = 0 , and a g = = 32 0 . ft s 2 . Thus, y v t at t i = + = + 1 2 144 0 16 0 2 2 ft ft s 2 . c h giving t fall s = 3 00 . . Her velocity just before impact is: v v gt f i = + = + ( )= 0 32 0 3 00 96 0 . . . ft s s ft s 2 c h . (b) While crushing the box, v i = 96 0 . ft s , v f = 0 , and y = = 18 0 1 50 . . in. ft. Therefore, a v v y f i = = ( ) = × 2 2 2 3 2 0 96 0 2 1 50 3 07 10 a f a f . . . ft s ft ft s 2 , or a = × 3 07 10 3 . ft s upward 2 . (c) Time to crush box: t y v y v v f i = = = ( ) + + 2 2 1 50 0 96 0 . . ft ft s or t = × 3 13 10 2 . s . P2.51 y t = 3 00 3 . : At t = 2 00 . s, y = = 3 00 2 00 24 0 3 . . . a f m and v dy dt t y = = = A 9 00 36 0 2 . . m s . If the helicopter releases a small mailbag at this time, the equation of motion of the mailbag is y y v t gt t t b bi i = + = + ( ) 1 2 24 0 36 0 1 2 9 80 2 2 . . . . Setting y b = 0, 0 24 0 36 0 4 90 2 = + . . . t t . Solving for t , (only positive values of
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