Question Number 178890 by peter frank last updated on 22/Oct/22
Commented by mr W last updated on 22/Oct/22
$${without}\:{this}\:{red}\:{word}\:{your}\:{question}\:{is} \\ $$$${not}\:{to}\:{understand}\:{properly}. \\ $$
Commented by mr W last updated on 22/Oct/22
Answered by mr W last updated on 22/Oct/22
![a=(dv/dt)=(ds/dt)×(dv/ds)=v(dv/ds) motion upwards: mv(dv/ds)=−mg−kmv^2 v(dv/ds)=−(g+kv^2 ) ((vdv)/(g+kv^2 ))=−ds ∫_v_0 ^v ((vdv)/( g+kv^2 ))=−∫_0 ^s ds ∫_v_0 ^v ((d(g+kv^2 ))/( g+kv^2 ))=−2k∫_0 ^s ds ln ((g+kv^2 )/(g+kv_0 ^2 ))=−2ks ⇒v^2 =(g/k)[(1+((kv_0 ^2 )/g))e^(−2ks) −1] at s=h_(max) : v=0 0=(g/k)[(1+((kv_0 ^2 )/g))e^(−2kh_(max) ) −1] ⇒e^(−2kh_(max) ) =(1/(1+((kv_0 ^2 )/g))) motion downwards: mv(dv/ds)=mg−kmv^2 ((vdv)/(g−kv^2 ))=ds ∫_0 ^v ((d(g−kv^2 ))/(g−kv^2 ))=−2k∫_0 ^s ds ln ((g−kv^2 )/g)=−2ks 1−((kv^2 )/g)=e^(−2ks) v^2 =(g/k)(1−e^(−2ks) ) when s→∞, v→v_c =terminal speed v_c ^2 =(g/k)(1−0)=(g/k) when at initial position: s=h_(max) v_1 ^2 =(g/k)(1−e^(−2kh_(max) ) ) v_1 ^2 =(g/k)(1−(1/(1+((kv_0 ^2 )/g)))) v_1 ^2 =((v_0 ^2 ×(g/k))/((g/k)+v_0 ^2 )) since (g/k)=v_c ^2 ⇒v_1 ^2 =((v_0 ^2 v_c ^2 )/(v_c ^2 +v_0 ^2 )) ⇒v_1 =((v_0 v_c )/( (√(v_0 ^2 +v_c ^2 )))) ✓](https://www.tinkutara.com/question/Q178896.png)
$${a}=\frac{{dv}}{{dt}}=\frac{{ds}}{{dt}}×\frac{{dv}}{{ds}}={v}\frac{{dv}}{{ds}} \\ $$$$ \\ $$$$\boldsymbol{{motion}}\:\boldsymbol{{upwards}}: \\ $$$${mv}\frac{{dv}}{{ds}}=−{mg}−{kmv}^{\mathrm{2}} \\ $$$${v}\frac{{dv}}{{ds}}=−\left({g}+{kv}^{\mathrm{2}} \right) \\ $$$$\frac{{vdv}}{{g}+{kv}^{\mathrm{2}} }=−{ds} \\ $$$$\int_{{v}_{\mathrm{0}} } ^{{v}} \frac{{vdv}}{\:{g}+{kv}^{\mathrm{2}} }=−\int_{\mathrm{0}} ^{{s}} {ds} \\ $$$$\int_{{v}_{\mathrm{0}} } ^{{v}} \frac{{d}\left({g}+{kv}^{\mathrm{2}} \right)}{\:{g}+{kv}^{\mathrm{2}} }=−\mathrm{2}{k}\int_{\mathrm{0}} ^{{s}} {ds} \\ $$$$\mathrm{ln}\:\frac{{g}+{kv}^{\mathrm{2}} }{{g}+{kv}_{\mathrm{0}} ^{\mathrm{2}} }=−\mathrm{2}{ks} \\ $$$$\Rightarrow{v}^{\mathrm{2}} =\frac{{g}}{{k}}\left[\left(\mathrm{1}+\frac{{kv}_{\mathrm{0}} ^{\mathrm{2}} }{{g}}\right){e}^{−\mathrm{2}{ks}} −\mathrm{1}\right] \\ $$$${at}\:{s}={h}_{{max}} :\:{v}=\mathrm{0} \\ $$$$\mathrm{0}=\frac{{g}}{{k}}\left[\left(\mathrm{1}+\frac{{kv}_{\mathrm{0}} ^{\mathrm{2}} }{{g}}\right){e}^{−\mathrm{2}{kh}_{{max}} } −\mathrm{1}\right] \\ $$$$\Rightarrow{e}^{−\mathrm{2}{kh}_{{max}} } =\frac{\mathrm{1}}{\mathrm{1}+\frac{{kv}_{\mathrm{0}} ^{\mathrm{2}} }{{g}}} \\ $$$$ \\ $$$$\boldsymbol{{motion}}\:\boldsymbol{{downwards}}: \\ $$$${mv}\frac{{dv}}{{ds}}={mg}−{kmv}^{\mathrm{2}} \\ $$$$\frac{{vdv}}{{g}−{kv}^{\mathrm{2}} }={ds} \\ $$$$\int_{\mathrm{0}} ^{{v}} \frac{{d}\left({g}−{kv}^{\mathrm{2}} \right)}{{g}−{kv}^{\mathrm{2}} }=−\mathrm{2}{k}\int_{\mathrm{0}} ^{{s}} {ds} \\ $$$$\mathrm{ln}\:\frac{{g}−{kv}^{\mathrm{2}} }{{g}}=−\mathrm{2}{ks} \\ $$$$\mathrm{1}−\frac{{kv}^{\mathrm{2}} }{{g}}={e}^{−\mathrm{2}{ks}} \\ $$$${v}^{\mathrm{2}} =\frac{{g}}{{k}}\left(\mathrm{1}−{e}^{−\mathrm{2}{ks}} \right) \\ $$$${when}\:{s}\rightarrow\infty,\:{v}\rightarrow{v}_{{c}} ={terminal}\:{speed} \\ $$$${v}_{{c}} ^{\mathrm{2}} =\frac{{g}}{{k}}\left(\mathrm{1}−\mathrm{0}\right)=\frac{{g}}{{k}} \\ $$$${when}\:{at}\:{initial}\:{position}:\:{s}={h}_{{max}} \\ $$$${v}_{\mathrm{1}} ^{\mathrm{2}} =\frac{{g}}{{k}}\left(\mathrm{1}−{e}^{−\mathrm{2}{kh}_{{max}} } \right) \\ $$$${v}_{\mathrm{1}} ^{\mathrm{2}} =\frac{{g}}{{k}}\left(\mathrm{1}−\frac{\mathrm{1}}{\mathrm{1}+\frac{{kv}_{\mathrm{0}} ^{\mathrm{2}} }{{g}}}\right) \\ $$$${v}_{\mathrm{1}} ^{\mathrm{2}} =\frac{{v}_{\mathrm{0}} ^{\mathrm{2}} ×\frac{{g}}{{k}}}{\frac{{g}}{{k}}+{v}_{\mathrm{0}} ^{\mathrm{2}} } \\ $$$${since}\:\frac{{g}}{{k}}={v}_{{c}} ^{\mathrm{2}} \\ $$$$\Rightarrow{v}_{\mathrm{1}} ^{\mathrm{2}} =\frac{{v}_{\mathrm{0}} ^{\mathrm{2}} {v}_{{c}} ^{\mathrm{2}} }{{v}_{{c}} ^{\mathrm{2}} +{v}_{\mathrm{0}} ^{\mathrm{2}} } \\ $$$$\Rightarrow{v}_{\mathrm{1}} =\frac{{v}_{\mathrm{0}} {v}_{{c}} }{\:\sqrt{{v}_{\mathrm{0}} ^{\mathrm{2}} +{v}_{{c}} ^{\mathrm{2}} }}\:\checkmark \\ $$
Commented by peter frank last updated on 22/Oct/22
$$\mathrm{thank}\:\mathrm{you} \\ $$
Commented by Tawa11 last updated on 23/Oct/22
$$\mathrm{Great}\:\mathrm{sir} \\ $$