Question Number 50330 by peter frank last updated on 15/Dec/18
Commented by peter frank last updated on 15/Dec/18
Answered by mr W last updated on 16/Dec/18
![a=mass of ball b=acceleration of ball ab=ag−R=ag−k^2 agv^2 ⇒b=g(1−k^2 v^2 ) since b=(dv/dt) ⇒(dv/dt)=g(1−k^2 v^2 ) ⇒(dv/(1−(kv)^2 ))=gdt ∫_0 ^v ((d(kv))/(1−(kv)^2 ))=kg∫_0 ^t dt ⇒(1/2)[ln ((1+kv)/(1−kv))]_0 ^v =kgt ⇒(1/2)ln ((1+kv)/(1−kv))=kgt ⇒((1+kv)/(1−kv))=e^(2kgt) =λ, say ⇒1+kv=λ−λkv ⇒(1+λ)kv=λ−1 ⇒v=(1/k)(((λ−1)/(λ+1))) ⇒v=(1/k)(((e^(2gkt) −1)/(e^(2gkt) +1))) =((e^(20kt) −1)/(k(e^(20kt) +1))) or ⇒v=(1/k)(1−(2/(1+e^(2gkt) ))) ⇒lim_(t→∞) v=(1/k)(1−(2/(1+∞)))=(1/k) ⇒limit velocity is (1/k) at t=(1/2): v=(1/(1/(10)))(1−(2/(1+e^(20×(1/(10))×(1/2)) )))=10(1−(2/(1+e)))≈4.621 m/s](Q50342.png)
$${a}={mass}\:{of}\:{ball} \\ $$$${b}={acceleration}\:{of}\:{ball} \\ $$$${ab}={ag}−{R}={ag}−{k}^{\mathrm{2}} {agv}^{\mathrm{2}} \\ $$$$\Rightarrow{b}={g}\left(\mathrm{1}−{k}^{\mathrm{2}} {v}^{\mathrm{2}} \right) \\ $$$${since}\:{b}=\frac{{dv}}{{dt}} \\ $$$$\Rightarrow\frac{{dv}}{{dt}}={g}\left(\mathrm{1}−{k}^{\mathrm{2}} {v}^{\mathrm{2}} \right) \\ $$$$\Rightarrow\frac{{dv}}{\mathrm{1}−\left({kv}\right)^{\mathrm{2}} }={gdt} \\ $$$$\int_{\mathrm{0}} ^{{v}} \frac{{d}\left({kv}\right)}{\mathrm{1}−\left({kv}\right)^{\mathrm{2}} }={kg}\int_{\mathrm{0}} ^{{t}} {dt} \\ $$$$\Rightarrow\frac{\mathrm{1}}{\mathrm{2}}\left[\mathrm{ln}\:\frac{\mathrm{1}+{kv}}{\mathrm{1}−{kv}}\right]_{\mathrm{0}} ^{{v}} ={kgt} \\ $$$$\Rightarrow\frac{\mathrm{1}}{\mathrm{2}}\mathrm{ln}\:\frac{\mathrm{1}+{kv}}{\mathrm{1}−{kv}}={kgt} \\ $$$$\Rightarrow\frac{\mathrm{1}+{kv}}{\mathrm{1}−{kv}}={e}^{\mathrm{2}{kgt}} =\lambda,\:{say} \\ $$$$\Rightarrow\mathrm{1}+{kv}=\lambda−\lambda{kv} \\ $$$$\Rightarrow\left(\mathrm{1}+\lambda\right){kv}=\lambda−\mathrm{1} \\ $$$$\Rightarrow{v}=\frac{\mathrm{1}}{{k}}\left(\frac{\lambda−\mathrm{1}}{\lambda+\mathrm{1}}\right) \\ $$$$\Rightarrow{v}=\frac{\mathrm{1}}{{k}}\left(\frac{{e}^{\mathrm{2}{gkt}} −\mathrm{1}}{{e}^{\mathrm{2}{gkt}} +\mathrm{1}}\right)\:=\frac{{e}^{\mathrm{20}{kt}} −\mathrm{1}}{{k}\left({e}^{\mathrm{20}{kt}} +\mathrm{1}\right)} \\ $$$${or} \\ $$$$\Rightarrow{v}=\frac{\mathrm{1}}{{k}}\left(\mathrm{1}−\frac{\mathrm{2}}{\mathrm{1}+{e}^{\mathrm{2}{gkt}} }\right) \\ $$$$ \\ $$$$\Rightarrow\underset{{t}\rightarrow\infty} {\mathrm{lim}}{v}=\frac{\mathrm{1}}{{k}}\left(\mathrm{1}−\frac{\mathrm{2}}{\mathrm{1}+\infty}\right)=\frac{\mathrm{1}}{{k}} \\ $$$$\Rightarrow{limit}\:{velocity}\:{is}\:\frac{\mathrm{1}}{{k}} \\ $$$$ \\ $$$${at}\:{t}=\frac{\mathrm{1}}{\mathrm{2}}: \\ $$$${v}=\frac{\mathrm{1}}{\frac{\mathrm{1}}{\mathrm{10}}}\left(\mathrm{1}−\frac{\mathrm{2}}{\mathrm{1}+{e}^{\mathrm{20}×\frac{\mathrm{1}}{\mathrm{10}}×\frac{\mathrm{1}}{\mathrm{2}}} }\right)=\mathrm{10}\left(\mathrm{1}−\frac{\mathrm{2}}{\mathrm{1}+{e}}\right)\approx\mathrm{4}.\mathrm{621}\:{m}/{s} \\ $$
Commented by peter frank last updated on 16/Dec/18
$${thank}\:{you}\:{sir} \\ $$