Question-78496

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Question Number 78496 by TawaTawa last updated on 18/Jan/20

Commented by mathmax by abdo last updated on 18/Jan/20

$${miss}\:{tawa}\:{are}\:{a}\:{theatcher}\:{or}\:{engineer}… \\ $$

Commented by TawaTawa last updated on 18/Jan/20

$$\mathrm{Engineer}\:\mathrm{sir}. \\ $$

Commented by mathmax by abdo last updated on 18/Jan/20

$${oh}\:{good}\:{job}\:{miss}\:{tawa}..\:{wich}\:{you}\:{goud}\:{luck}\:… \\ $$

Commented by TawaTawa last updated on 19/Jan/20

$$\mathrm{Thank}\:\mathrm{you}\:\mathrm{sir},\:\:\mathrm{i}\:\mathrm{appreciate}. \\ $$

Answered by mr W last updated on 18/Jan/20

ma=−mg−Kmv a=−g−Kv v(dv/dh)=−g−Kv ((vdv)/(g+Kv))=−dh ∫_u ^0 ((vdv)/(g+Kv))=−∫_0 ^H dh [(v/K)−((g ln (g+Kv))/K^2 )]_u ^0 =−H −((g ln g)/K^2 )−(u/K)+((g ln (g+Ku))/K^2 )=−H ⇒H=(u/K)−(g/K^2 )ln (1+((Ku)/g))=z_(max)

$${ma}=−{mg}−{Kmv} \\ $$$${a}=−{g}−{Kv} \\ $$$${v}\frac{{dv}}{{dh}}=−{g}−{Kv} \\ $$$$\frac{{vdv}}{{g}+{Kv}}=−{dh} \\ $$$$\int_{{u}} ^{\mathrm{0}} \frac{{vdv}}{{g}+{Kv}}=−\int_{\mathrm{0}} ^{{H}} {dh} \\ $$$$\left[\frac{{v}}{{K}}−\frac{{g}\:\mathrm{ln}\:\left({g}+{Kv}\right)}{{K}^{\mathrm{2}} }\right]_{{u}} ^{\mathrm{0}} =−{H} \\ $$$$−\frac{{g}\:\mathrm{ln}\:{g}}{{K}^{\mathrm{2}} }−\frac{{u}}{{K}}+\frac{{g}\:\mathrm{ln}\:\left({g}+{Ku}\right)}{{K}^{\mathrm{2}} }=−{H} \\ $$$$\Rightarrow{H}=\frac{{u}}{{K}}−\frac{{g}}{{K}^{\mathrm{2}} }\mathrm{ln}\:\left(\mathrm{1}+\frac{{Ku}}{{g}}\right)={z}_{{max}} \\ $$

Commented by peter frank last updated on 18/Jan/20