Chapter 241P2.50Take downward as the positive ydirection.(a)While the woman was in free fall,∆y=144 ft, vi=0 , and ag==32 0. fts2.Thus, ∆yvt atti=+→12144016 022ftft s2.chgiving tfalls=300.. Her velocity justbefore impact is:vvgtfi=+=+()=0320960...ft ssft s2.(b)While crushing the box, vi=96 0f,vf=0, and ∆y18 01 50in.ft. Therefore,ay=−=−()=−×23209602150307 10∆af...ft sftft s2, or a=×3ftsupward2.(c)Time to crush box: ∆∆∆tyvy=++260..ftft sor ∆t−313 102s.P2.51yt=3.: At t=200s,y3 00 2 0024 03.afm andvdydtty=A9003602ms.If the helicopter releases a small mailbag at this time, the equation of motion of the mailbag isyyvtgtttbbii−−1224 036 012980..Setting yb=0,024 036 04 902−...tt.Solving for t, (only positive values of tcount), t=796s.*P2.52Consider the last 30 m of fall. We find its speed 30 m above the ground:
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