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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-spee




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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