Question Solved1 Answer The motion of a parachutist free-falling from rest from a stationary helicopter is given by the differential equation dy v=9.8 -0.002v2 dx where x m is her distance below the helicopter and v ms' is her velocity. Solve the differential equation to show that y= = 70(1-e e 0.004x))/2 [6]

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Transcribed Image Text: The motion of a parachutist free-falling from rest from a stationary helicopter is given by the differential equation dy v=9.8 -0.002v2 dx where x m is her distance below the helicopter and v ms' is her velocity. Solve the differential equation to show that y= = 70(1-e e 0.004x))/2 [6]
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Transcribed Image Text: The motion of a parachutist free-falling from rest from a stationary helicopter is given by the differential equation dy v=9.8 -0.002v2 dx where x m is her distance below the helicopter and v ms' is her velocity. Solve the differential equation to show that y= = 70(1-e e 0.004x))/2 [6]
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V(dV)/(dx)=(9:8-0.002V^(2))int(vdv)/(9.8-0.002v^(2))=int dx(1)/(-2xx0.002)(d(9.8-0.002y^(2)))/(9.8-0.002y^(2))=x+ln C-(1)/(1//250)ln(9.8-0.002y^(2))=x+ln C-250 ln(9.8-0.002x^(2))=x+ln cln(9.8-0.002v^(2))=(-1)/(250)(x+ln C)9.8-0.002V^ ... See the full answer