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Question Number 18264 by Tinkutara last updated on 17/Jul/17

A sky diver of mass m drops out with an initial velocity v_0 = 0. Find the law by which the sky diver′s speed varies before the parachute is opened if the drag is proportional to the sky diver′s speed. Also solve the problem when the sky diver′s initial velocity has horizontal component v_0 and vertical component zero.

$$\mathrm{A}\:\mathrm{sky}\:\mathrm{diver}\:\mathrm{of}\:\mathrm{mass}\:{m}\:\mathrm{drops}\:\mathrm{out}\:\mathrm{with} \\ $$$$\mathrm{an}\:\mathrm{initial}\:\mathrm{velocity}\:{v}_{\mathrm{0}} \:=\:\mathrm{0}.\:\mathrm{Find}\:\mathrm{the}\:\mathrm{law} \\ $$$$\mathrm{by}\:\mathrm{which}\:\mathrm{the}\:\mathrm{sky}\:\mathrm{diver}'\mathrm{s}\:\mathrm{speed}\:\mathrm{varies} \\ $$$$\mathrm{before}\:\mathrm{the}\:\mathrm{parachute}\:\mathrm{is}\:\mathrm{opened}\:\mathrm{if}\:\mathrm{the} \\ $$$$\mathrm{drag}\:\mathrm{is}\:\mathrm{proportional}\:\mathrm{to}\:\mathrm{the}\:\mathrm{sky}\:\mathrm{diver}'\mathrm{s} \\ $$$$\mathrm{speed}.\:\mathrm{Also}\:\mathrm{solve}\:\mathrm{the}\:\mathrm{problem}\:\mathrm{when}\:\mathrm{the} \\ $$$$\mathrm{sky}\:\mathrm{diver}'\mathrm{s}\:\mathrm{initial}\:\mathrm{velocity}\:\mathrm{has}\:\mathrm{horizontal} \\ $$$$\mathrm{component}\:{v}_{\mathrm{0}} \:\mathrm{and}\:\mathrm{vertical}\:\mathrm{component} \\ $$$$\mathrm{zero}. \\ $$

Answered by ajfour last updated on 16/Aug/17

Commented by ajfour last updated on 16/Aug/17

let drag force be F^� =−kv^� As ΣF_y =((mdv_y )/dt) −kv_y −mg=m(dv_y /dt) ∫_0 ^( v_y ) ((kdv_y )/(kv_y +mg))=−(k/m)∫_0 ^( t) dt ⇒ ln (((kv_y +mg)/(mg)))=−((kt)/m) kv_y +mg=mge^(−kt/m) ⇒ v_y =−((mg)/k)(1−e^(−kt/m) ) ...(i) Further As ΣF_x =m(dv_x /dt) ⇒ −kv_x =((mdv_x )/dt) ∫_( v_0 ) ^( v_x ) (dv_x /v_x )=−((kt)/m)∫_0 ^( t) dt ⇒ ln (v_x /v_0 )=e^(−kt/m) ⇒ v_x =v_0 e^(−kt/m) ....(ii) As v^2 =v_x ^2 +v_y ^2 , so from (i)&(ii) v={v_0 ^2 e^(−2kt/m) +(((mg)/k))^2 [1−e^(−kt/m) ]^2 }^(1/2) If v_0 =0, v=((mg)/k)(1−e^(−kt/m) ) .

$$\:\:\:{let}\:{drag}\:{force}\:{be}\:\bar {{F}}=−{k}\bar {{v}} \\ $$$${As}\:\:\:\:\Sigma{F}_{{y}} =\frac{{mdv}_{{y}} }{{dt}} \\ $$$$\:\:\:\:\:−{kv}_{{y}} −{mg}={m}\frac{{dv}_{{y}} }{{dt}} \\ $$$$\:\:\:\:\int_{\mathrm{0}} ^{\:\:\boldsymbol{{v}}_{\boldsymbol{{y}}} } \frac{{kdv}_{{y}} }{{kv}_{{y}} +{mg}}=−\frac{{k}}{{m}}\int_{\mathrm{0}} ^{\:\:{t}} {dt} \\ $$$$\Rightarrow\:\:\:\:\mathrm{ln}\:\left(\frac{{kv}_{{y}} +{mg}}{{mg}}\right)=−\frac{{kt}}{{m}} \\ $$$$\:\:\:\:\:\:{kv}_{{y}} +{mg}={mge}^{−{kt}/{m}} \\ $$$$\Rightarrow\:\:\:\:\:\:\:{v}_{{y}} =−\frac{{mg}}{{k}}\left(\mathrm{1}−{e}^{−{kt}/{m}} \right)\:\:\:...\left(\boldsymbol{{i}}\right) \\ $$$${Further}\:{As}\:\:\Sigma{F}_{{x}} ={m}\frac{{dv}_{{x}} }{{dt}} \\ $$$$\Rightarrow\:\:\:\:\:−{kv}_{{x}} =\frac{{mdv}_{{x}} }{{dt}} \\ $$$$\:\:\:\:\:\:\int_{\:{v}_{\mathrm{0}} } ^{\:\:\boldsymbol{{v}}_{\boldsymbol{{x}}} } \:\frac{{dv}_{{x}} }{{v}_{{x}} }=−\frac{{kt}}{{m}}\int_{\mathrm{0}} ^{\:\:{t}} {dt} \\ $$$$\Rightarrow\:\:\:\:\mathrm{ln}\:\frac{{v}_{{x}} }{{v}_{\mathrm{0}} }={e}^{−{kt}/{m}} \\ $$$$\Rightarrow\:\:\:\:\:\:{v}_{{x}} ={v}_{\mathrm{0}} {e}^{−{kt}/{m}} \:\:\:\:....\left(\boldsymbol{{ii}}\right) \\ $$$$\:{As}\:\:\:{v}^{\mathrm{2}} ={v}_{{x}} ^{\mathrm{2}} +{v}_{{y}} ^{\mathrm{2}} \:\:\:,\:{so}\:{from}\:\left({i}\right)\&\left({ii}\right) \\ $$$$\:\boldsymbol{{v}}=\left\{\boldsymbol{{v}}_{\mathrm{0}} ^{\mathrm{2}} \boldsymbol{{e}}^{−\mathrm{2}\boldsymbol{{kt}}/\boldsymbol{{m}}} +\left(\frac{\boldsymbol{{mg}}}{\boldsymbol{{k}}}\right)^{\mathrm{2}} \left[\mathrm{1}−\boldsymbol{{e}}^{−\boldsymbol{{kt}}/\boldsymbol{{m}}} \right]^{\mathrm{2}} \:\right\}^{\mathrm{1}/\mathrm{2}} \\ $$$${If}\:{v}_{\mathrm{0}} =\mathrm{0},\:\:\:{v}=\frac{{mg}}{{k}}\left(\mathrm{1}−{e}^{−{kt}/{m}} \right)\:. \\ $$

Commented by Tinkutara last updated on 16/Aug/17

Thank you very much Sir!

$$\mathrm{Thank}\:\mathrm{you}\:\mathrm{very}\:\mathrm{much}\:\mathrm{Sir}! \\ $$

Commented by Tinkutara last updated on 30/Aug/17

But the body is accelerating downwards, so it should be mg − (−kv) = m(dv_x /dt) but why you have written −kv−mg =m(dv_x /dt)?

$$\mathrm{But}\:\mathrm{the}\:\mathrm{body}\:\mathrm{is}\:\mathrm{accelerating}\:\mathrm{downwards}, \\ $$$$\mathrm{so}\:\mathrm{it}\:\mathrm{should}\:\mathrm{be}\:{mg}\:−\:\left(−{kv}\right)\:=\:{m}\frac{{dv}_{{x}} }{{dt}} \\ $$$$\mathrm{but}\:\mathrm{why}\:\mathrm{you}\:\mathrm{have}\:\mathrm{written}\:−{kv}−{mg} \\ $$$$={m}\frac{{dv}_{{x}} }{{dt}}? \\ $$

Commented by ajfour last updated on 30/Aug/17

do notice that v_y < 0 for the situation marked in diagram the eqation agrees to both when body is going up or coming down. drag force and v_y are oppositely directed , force of gravity downwards. i have solved for the two dimensional case right from the beginning; ΣF_y =−mg−kv_y =m(dv_y /dt) i have written. when v_y <0 drag force component −kv_y >0 .

$${do}\:{notice}\:{that}\:\boldsymbol{{v}}_{\boldsymbol{{y}}} \:<\:\mathrm{0}\:{for}\:{the} \\ $$$${situation}\:{marked}\:{in}\:{diagram} \\ $$$${the}\:{eqation}\:{agrees}\:{to}\:{both}\:{when} \\ $$$${body}\:{is}\:{going}\:{up}\:{or}\:{coming}\:{down}. \\ $$$${drag}\:{force}\:{and}\:{v}_{{y}} \:{are}\:{oppositely} \\ $$$${directed}\:,\:{force}\:{of}\:{gravity} \\ $$$${downwards}.\:{i}\:{have}\:{solved}\: \\ $$$${for}\:{the}\:{two}\:{dimensional}\:{case} \\ $$$${right}\:{from}\:{the}\:{beginning}; \\ $$$$\Sigma{F}_{{y}} =−{mg}−{kv}_{{y}} ={m}\frac{{dv}_{{y}} }{{dt}} \\ $$$${i}\:{have}\:{written}.\:{when}\:{v}_{{y}} <\mathrm{0} \\ $$$${drag}\:{force}\:{component}\:−{kv}_{{y}} >\mathrm{0}\:. \\ $$

Commented by Tinkutara last updated on 31/Aug/17

OK. Thanks!

$$\mathrm{OK}.\:\mathrm{Thanks}! \\ $$

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