Question-209161

Question Number 209161 by Spillover last updated on 02/Jul/24

Commented by klipto last updated on 03/Jul/24

F_(net) =−(F_g +P) ma=−(mg+0.2v^2 ) a=−(g+(v^2 /(5m))) (dv/dt)=−(g+(v^2 /(5m))) also a=v×(dv/dh) ∴−(g+(v^2 /(5m)))=((vdv)/dh) ((vdv)/(g+(v^2 /(5m))))=−dh (v/(g+(v^2 /(10))))dv=−dh (v/(g+0.1v^2 ))dv=−dh ∫_u ^v (v/(g+0.1v^2 ))dv=−∫dh , [u=g+0.1v^2 ),dv=(du/(0.2v)) ∫_u ^v (v/u)×(du/(0.2v))=−h (1/(0.2))∫_(10) ^0 (du/u)=−h 5inu∣_(10) ^0 =−h 5[in(g+0.1v^2 )∣^0 −in(g+0.1v^2 )∣^(10) =−h 5[in(10)−in(10+((100)/(10)))]=5[in10−in20]=−h −5ln2=−h h=3.47 klipto−quanta

$$ \\ $$$$\boldsymbol{\mathrm{F}}_{\boldsymbol{\mathrm{net}}} =−\left(\boldsymbol{\mathrm{F}}_{\mathrm{g}} +\boldsymbol{\mathrm{P}}\right) \\ $$$$\boldsymbol{\mathrm{ma}}=−\left(\boldsymbol{\mathrm{mg}}+\mathrm{0}.\mathrm{2}\boldsymbol{\mathrm{v}}^{\mathrm{2}} \right) \\ $$$$\boldsymbol{\mathrm{a}}=−\left(\boldsymbol{\mathrm{g}}+\frac{\boldsymbol{\mathrm{v}}^{\mathrm{2}} }{\mathrm{5}\boldsymbol{\mathrm{m}}}\right) \\ $$$$\frac{\boldsymbol{\mathrm{dv}}}{\boldsymbol{\mathrm{dt}}}=−\left(\boldsymbol{\mathrm{g}}+\frac{\boldsymbol{\mathrm{v}}^{\mathrm{2}} }{\mathrm{5}\boldsymbol{\mathrm{m}}}\right) \\ $$$$\boldsymbol{\mathrm{also}} \\ $$$$\boldsymbol{\mathrm{a}}=\boldsymbol{\mathrm{v}}×\frac{\boldsymbol{\mathrm{dv}}}{\boldsymbol{\mathrm{dh}}} \\ $$$$\therefore−\left(\boldsymbol{\mathrm{g}}+\frac{\boldsymbol{\mathrm{v}}^{\mathrm{2}} }{\mathrm{5}\boldsymbol{\mathrm{m}}}\right)=\frac{\boldsymbol{\mathrm{vdv}}}{\boldsymbol{\mathrm{dh}}} \\ $$$$\frac{\boldsymbol{\mathrm{vdv}}}{\boldsymbol{\mathrm{g}}+\frac{\boldsymbol{\mathrm{v}}^{\mathrm{2}} }{\mathrm{5}\boldsymbol{\mathrm{m}}}}=−\boldsymbol{\mathrm{dh}} \\ $$$$\frac{\boldsymbol{\mathrm{v}}}{\boldsymbol{\mathrm{g}}+\frac{\boldsymbol{\mathrm{v}}^{\mathrm{2}} }{\mathrm{10}}}\boldsymbol{\mathrm{dv}}=−\boldsymbol{\mathrm{dh}} \\ $$$$\frac{\boldsymbol{\mathrm{v}}}{\boldsymbol{\mathrm{g}}+\mathrm{0}.\mathrm{1}\boldsymbol{\mathrm{v}}^{\mathrm{2}} }\boldsymbol{\mathrm{dv}}=−\boldsymbol{\mathrm{dh}} \\ $$$$\int_{\boldsymbol{\mathrm{u}}} ^{\boldsymbol{\mathrm{v}}} \frac{\boldsymbol{\mathrm{v}}}{\boldsymbol{\mathrm{g}}+\mathrm{0}.\mathrm{1}\boldsymbol{\mathrm{v}}^{\mathrm{2}} }\boldsymbol{\mathrm{dv}}=−\int\boldsymbol{\mathrm{dh}}\:,\:\left[\boldsymbol{\mathrm{u}}=\boldsymbol{\mathrm{g}}+\mathrm{0}.\mathrm{1}\boldsymbol{\mathrm{v}}^{\mathrm{2}} \right),\boldsymbol{\mathrm{dv}}=\frac{\boldsymbol{\mathrm{du}}}{\mathrm{0}.\mathrm{2}\boldsymbol{\mathrm{v}}} \\ $$$$\int_{\boldsymbol{\mathrm{u}}} ^{\boldsymbol{\mathrm{v}}} \frac{\boldsymbol{\mathrm{v}}}{\boldsymbol{\mathrm{u}}}×\frac{\boldsymbol{\mathrm{du}}}{\mathrm{0}.\mathrm{2}\boldsymbol{\mathrm{v}}}=−\boldsymbol{\mathrm{h}} \\ $$$$\frac{\mathrm{1}}{\mathrm{0}.\mathrm{2}}\int_{\mathrm{10}} ^{\mathrm{0}} \frac{\boldsymbol{\mathrm{du}}}{\boldsymbol{\mathrm{u}}}=−\boldsymbol{\mathrm{h}} \\ $$$$\mathrm{5}\boldsymbol{\mathrm{inu}}\mid_{\mathrm{10}} ^{\mathrm{0}} =−\boldsymbol{\mathrm{h}} \\ $$$$\mathrm{5}\left[\boldsymbol{\mathrm{in}}\left(\boldsymbol{\mathrm{g}}+\mathrm{0}.\mathrm{1}\boldsymbol{\mathrm{v}}^{\mathrm{2}} \right)\mid^{\mathrm{0}} −\boldsymbol{\mathrm{in}}\left(\boldsymbol{\mathrm{g}}+\mathrm{0}.\mathrm{1}\boldsymbol{\mathrm{v}}^{\mathrm{2}} \right)\mid^{\mathrm{10}} =−\boldsymbol{\mathrm{h}}\right. \\ $$$$\mathrm{5}\left[\boldsymbol{\mathrm{in}}\left(\mathrm{10}\right)−\boldsymbol{\mathrm{in}}\left(\mathrm{10}+\frac{\mathrm{100}}{\mathrm{10}}\right)\right]=\mathrm{5}\left[\boldsymbol{\mathrm{in}}\mathrm{10}−\boldsymbol{\mathrm{in}}\mathrm{20}\right]=−\boldsymbol{\mathrm{h}} \\ $$$$−\mathrm{5}\boldsymbol{\mathrm{ln}}\mathrm{2}=−\boldsymbol{\mathrm{h}} \\ $$$$\boldsymbol{\mathrm{h}}=\mathrm{3}.\mathrm{47} \\ $$$$\boldsymbol{\mathrm{klipto}}−\boldsymbol{\mathrm{quanta}} \\ $$

Commented by Spillover last updated on 03/Jul/24

$${thank}\:{you} \\ $$

Answered by mr W last updated on 02/Jul/24

ma=−mg−(v^2 /5) a=−g−(v^2 /(5m))=(dv/dt)=v(dv/dh) −(g+(v^2 /(5m)))=v(dv/dh) ((vdv)/(g+(v^2 /(5m))))=−dh ∫((v^2 dv)/(g+(v^2 /(5m))))=−2∫dh ln (g+(v^2 /(5m)))∣_u ^v =−((2h)/(5m)) ln ((g+(v^2 /(5m)))/(g+(u^2 /(5m))))=−((2h)/(5m)) at h_(max) : v=0 h_(max) =((5m)/2)ln (1+(u^2 /(5mg))) =((5×2)/2)ln (1+((10^2 )/(5×2×10)))=3.47 m

$${ma}=−{mg}−\frac{{v}^{\mathrm{2}} }{\mathrm{5}} \\ $$$${a}=−{g}−\frac{{v}^{\mathrm{2}} }{\mathrm{5}{m}}=\frac{{dv}}{{dt}}={v}\frac{{dv}}{{dh}} \\ $$$$−\left({g}+\frac{{v}^{\mathrm{2}} }{\mathrm{5}{m}}\right)={v}\frac{{dv}}{{dh}} \\ $$$$\frac{{vdv}}{{g}+\frac{{v}^{\mathrm{2}} }{\mathrm{5}{m}}}=−{dh} \\ $$$$\int\frac{{v}^{\mathrm{2}} {dv}}{{g}+\frac{{v}^{\mathrm{2}} }{\mathrm{5}{m}}}=−\mathrm{2}\int{dh} \\ $$$$\mathrm{ln}\:\left({g}+\frac{{v}^{\mathrm{2}} }{\mathrm{5}{m}}\right)\mid_{{u}} ^{{v}} =−\frac{\mathrm{2}{h}}{\mathrm{5}{m}} \\ $$$$\mathrm{ln}\:\frac{{g}+\frac{{v}^{\mathrm{2}} }{\mathrm{5}{m}}}{{g}+\frac{{u}^{\mathrm{2}} }{\mathrm{5}{m}}}=−\frac{\mathrm{2}{h}}{\mathrm{5}{m}} \\ $$$$ \\ $$$${at}\:{h}_{{max}} :\:{v}=\mathrm{0} \\ $$$${h}_{{max}} =\frac{\mathrm{5}{m}}{\mathrm{2}}\mathrm{ln}\:\left(\mathrm{1}+\frac{{u}^{\mathrm{2}} }{\mathrm{5}{mg}}\right) \\ $$$$\:\:\:\:\:\:\:\:\:\:=\frac{\mathrm{5}×\mathrm{2}}{\mathrm{2}}\mathrm{ln}\:\left(\mathrm{1}+\frac{\mathrm{10}^{\mathrm{2}} }{\mathrm{5}×\mathrm{2}×\mathrm{10}}\right)=\mathrm{3}.\mathrm{47}\:{m} \\ $$

Commented by Spillover last updated on 02/Jul/24