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 ag == 32 0 . f ts 2 . Thus, yv t a t t i =+ 1 2 144 0 16 0 22 ft ft s 2 . ch giving t fall s = 300 . . Her velocity just before impact is: vv g t fi =+=+ ( ) = 0 320 960 .. . ft s s ft s 2 . (b) While crushing the box, v i = 96 0 f , v f = 0, and y 18 0 1 50 in. ft. Therefore, a y = = () = × 2 3 2 09 6 0 2150 307 10 af . . . ft s ft ft s 2 , or a 3 f t s u p w a r d 2 . (c) Time to crush box: ∆∆ t y v y = + + 2 6 0 . . ft ft s or t 313 10 2 s . P2.51 yt = 3 . : At t = 200 s , y 3 00 2 00 24 0 3 . a f m and v dy dt t y = A 900 360 2 m s . If the helicopter releases a small mailbag at this time, the equation of motion of the mailbag is yyv t g t t t bb ii 1 2 24 0 36 0 1 2 980 . . Setting y b = 0, 0 24 0 36 0 4 90 2 ... tt . Solving for t , (only positive values of t count), t = 796 s . *P2.52 Consider the last 30 m of fall. We find its speed 30 m above the ground:
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