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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

$$a=massofball$$$$b=accelerationofball$$$$ab=ag-R=ag-k^2{agv}^2$$$$\Rightarrow b=g(1-k^2{v}^2)$$$$sinceb=\frac{dv}{dt}$$$$\Rightarrow \frac{dv}{dt}=g(1-k^2{v}^2)$$$$\Rightarrow \frac{dv}{1-{\left(kv\right)}^2}=gdt$$$${\int }_0^v\frac{d\left(kv\right)}{1-{\left(kv\right)}^2}=kg{\int }_0^{t}dt$$$$\Rightarrow \frac{1}{2}{\left[\ln \frac{1+kv}{1-kv}\right]}_0^v=kgt$$$$\Rightarrow \frac{1}{2}\ln \frac{1+kv}{1-kv}=kgt$$$$\Rightarrow \frac{1+kv}{1-kv}=e^{2kgt}=\lambda ,say$$$$\Rightarrow 1+kv=\lambda -\lambda kv$$$$\Rightarrow (1+\lambda )kv=\lambda -1$$$$\Rightarrow v=\frac{1}{k}\left(\frac{\lambda -1}{\lambda +1}\right)$$$$\Rightarrow v=\frac{1}{k}\left(\frac{e^{2gkt}-1}{e^{2gkt}+1}\right)=\frac{e^{20kt}-1}{k(e^{20kt}+1)}$$$$or$$$$\Rightarrow v=\frac{1}{k}(1-\frac{2}{1+e^{2gkt}})$$$$$$$$\Rightarrow \underset{t\to \mathrm{\infty }}{\lim }v=\frac{1}{k}(1-\frac{2}{1+\mathrm{\infty }})=\frac{1}{k}$$$$\Rightarrow limitvelocityis\frac{1}{k}$$$$$$$$att=\frac{1}{2}:$$$$v=\frac{1}{\frac{1}{10}}(1-\frac{2}{1+e^{20\times \frac{1}{10}\times \frac{1}{2}}})=10(1-\frac{2}{1+e})\approx 4.621m/s$$

Commented by peter frank last updated on 16/Dec/18

$$thankyousir$$

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