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Question Number 178782 by peter frank last updated on 21/Oct/22

Answered by mr W last updated on 22/Oct/22

$$m\frac{dv}{dt}=mg\sin \theta -{kmv}^2$$$$dt=\frac{dv}{g\sin \theta -{kv}^2}$$$${\int }_0^{t}dt={\int }_0^v\frac{dv}{g\sin \theta -{kv}^2}$$$${\int }_0^{t}dt={\int }_0^v\frac{d\left(kv\right)}{{\left(\sqrt{kg\sin \theta }\right)}^2-{\left(kv\right)}^2}$$$$t=\frac{1}{2\sqrt{kg\sin \theta }}{\left[\ln \frac{\sqrt{kg\sin \theta }+kv}{\sqrt{kg\sin \theta }-kv}\right]}_0^v$$$$2\sqrt{kg\sin \theta }t=\ln \frac{\sqrt{kg\sin \theta }+kv}{\sqrt{kg\sin \theta }-kv}$$$$e^{2\sqrt{kg\sin \theta }t}=\frac{\sqrt{kg\sin \theta }+kv}{\sqrt{kg\sin \theta }-kv}$$$$let\lambda =\sqrt{kg\sin \theta }$$$$e^{2\lambda t}=\frac{2\lambda }{\lambda -kv}-1$$$$\frac{\lambda }{k}(1-\frac{2}{1+e^{2\lambda t}})=v=\frac{ds}{dt}$$$$\frac{\lambda }{k}(1-\frac{2}{1+e^{2\lambda t}})dt=ds$$$$\frac{\lambda }{k}{\int }_0^t(1-\frac{2}{1+e^{2\lambda t}})dt={\int }_0^{d}ds$$$$t-{\int }_0^t\frac{2}{1+e^{2\lambda t}}dt=\frac{kd}{\lambda }$$$$t-\frac{1}{\lambda }{\int }_0^t\frac{d\left(2\lambda t\right)}{1+e^{2\lambda t}}=\frac{kd}{\lambda }$$$$t-\frac{1}{\lambda }{[2\lambda t-\ln (1+e^{2\lambda t})]}_0^t=\frac{kd}{\lambda }$$$$t-\frac{1}{\lambda }[2\lambda t-\ln (1+e^{2\lambda t})+\ln 2]=\frac{kd}{\lambda }$$$$t-\frac{1}{\lambda }\ln \frac{2e^{2\lambda t}}{1+e^{2\lambda t}}=\frac{kd}{\lambda }$$$$\ln e^{\lambda t}-\ln \frac{2e^{2\lambda t}}{1+e^{2\lambda t}}=kd$$$$\ln \frac{(1+e^{2\lambda t})}{2e^{\lambda t}}=kd$$$$\frac{1+e^{2\lambda t}}{2e^{\lambda t}}=e^{kd}$$$${\left(e^{\lambda t}\right)}^2-2e^{kd}{e}^{\lambda t}+1=0$$$$e^{\lambda t}=e^{kd}+\sqrt{e^{2kd}-1}$$$$\lambda t=\ln (e^{kd}+\sqrt{e^{2kd}-1})={\cosh }^{-1}\left(e^{kd}\right)$$$$\Rightarrow t=\frac{{\cosh }^{-1}\left(e^{kd}\right)}{\lambda }$$$$\Rightarrow t=\frac{{\cosh }^{-1}\left(e^{kd}\right)}{\sqrt{kg\sin \theta }}✓$$

Commented by mr W last updated on 22/Oct/22

Commented by Tawa11 last updated on 22/Oct/22

$$\mathrm{Great}\mathrm{sir}$$

Commented by Spillover last updated on 22/Oct/22

$$\mathrm{Excellent}\mathrm{solution}\mathrm{Mr}\mathrm{W}$$

Commented by peter frank last updated on 22/Oct/22

$$\mathrm{thanks}$$

Answered by Spillover last updated on 22/Oct/22

$$\mathrm{F}=\mathrm{ma}$$$$-{\mathrm{kmv}}^2+\mathrm{mgsin}\theta =\mathrm{ma}$$$$\mathrm{gsin}\theta -{\mathrm{kv}}^2=\mathrm{a}$$$$\frac{\mathrm{dv}}{\mathrm{dt}}=\mathrm{gsin}\theta -{\mathrm{kv}}^2=\mathrm{k}[\frac{\mathrm{g}}{\mathrm{k}}\sin \theta -{\mathrm{v}}^2]$$$$\mathrm{dt}=\int \frac{\mathrm{dv}}{\mathrm{k}[\frac{\mathrm{g}}{\mathrm{k}}\sin \theta -{\mathrm{v}}^2]}=\frac{1}{\mathrm{k}}\int \frac{\mathrm{dv}}{\frac{\mathrm{g}}{\mathrm{k}}\sin \theta -{\mathrm{v}}^2}$$$$\mathrm{dt}=\frac{1}{\mathrm{k}}\int \frac{\mathrm{dv}}{\frac{\mathrm{g}}{\mathrm{k}}\sin \theta -{\mathrm{v}}^2}=\frac{1}{\mathrm{k}}\int \frac{\mathrm{dv}}{\sqrt{{\left(\frac{\mathrm{g}}{\mathrm{k}}\sin \theta \right)}^2}-{\mathrm{v}}^2}$$$$\mathrm{let}\beta =\sqrt{\frac{\mathrm{g}}{\mathrm{k}}\sin \theta }$$$$\mathrm{dt}=\frac{1}{\mathrm{k}}\int \frac{\mathrm{dv}}{\sqrt{{\left(\frac{\mathrm{g}}{\mathrm{k}}\sin \theta \right)}^2}-{\mathrm{v}}^2}=\frac{1}{\mathrm{k}}\int \frac{\mathrm{dv}}{{\beta }^2-{\mathrm{v}}^2}$$$$\mathrm{t}=\frac{1}{\mathrm{k}}.\frac{1}{\beta }\tanh {}^{-1}\left(\frac{\mathrm{v}}{\beta }\right)+\mathrm{A}$$$$\mathrm{t}=0\mathrm{x}=0\mathrm{v}=0$$$$\frac{\mathrm{v}}{\beta }=\tanh \left(\mathrm{kt}\beta \right)\mathrm{v}=\beta \tanh \left(\mathrm{kt}\beta \right)$$$$\mathrm{v}=\beta \tanh \left(\mathrm{kt}\beta \right)\frac{\mathrm{dx}}{\mathrm{dt}}=\beta \tanh \left(\mathrm{kt}\beta \right)\mathrm{dx}=\beta \tanh \left(\mathrm{kt}\beta \right)\mathrm{dt}$$$${\int }_0^{\mathrm{d}}\mathrm{dx}=\int \beta \tanh \left(\mathrm{kt}\beta \right)\mathrm{dt}$$$$\mathrm{d}=\beta \ln \frac{[\cosh \left(\mathrm{kt}\beta \right)}{\mathrm{k}\beta }$$$$\mathrm{d}=\frac{1}{\mathrm{k}}\ln \cosh \left(\mathrm{kt}\beta \right)$$$$\mathrm{kd}=\ln \cosh \left(\mathrm{kt}\beta \right)$$$${\mathrm{e}}^{\mathrm{kd}}=\cosh \left(\mathrm{kt}\beta \right)$$$$\cosh {}^{-1}\left({\mathrm{e}}^{\mathrm{kd}}\right)=\mathrm{kt}\beta$$$$\mathrm{t}=\frac{1}{\mathrm{k}\beta }\cosh {}^{-1}\left({\mathrm{e}}^{\mathrm{kd}}\right)$$$$\beta =\sqrt{\frac{\mathrm{g}}{\mathrm{k}}\sin \theta }$$$$\mathrm{t}=\frac{1}{\mathrm{k}.\sqrt{\frac{\mathrm{g}}{\mathrm{k}}\sin \theta }}.\cosh {}^{-1}\left({\mathrm{e}}^{\mathrm{kd}}\right)$$$$\mathrm{t}=\frac{\cosh {}^{-1}\left({\mathrm{e}}^{\mathrm{kd}}\right)}{\sqrt{\mathrm{gksin}\theta }}$$$$$$$$$$

Commented by peter frank last updated on 22/Oct/22

$$\mathrm{great}\mathrm{sir}$$

Commented by mr W last updated on 22/Oct/22

$$wecanseeyouareafanofhyperbolic$$$$functions.{that}^{\prime }sgreat!$$

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