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Question Number 137618 by I want to learn more last updated on 04/Apr/21

Commented by I want to learn more last updated on 04/Apr/21

$$\mathrm{Find}\mathrm{the}\mathrm{range}\mathrm{of}\mathrm{nth}\mathrm{trajectory}.$$$$\mathrm{Here}\mathrm{they}\mathrm{wrote}.\frac{{\mathrm{e}}^{\mathrm{n}-1}{\mathrm{u}}^2\sin \left(2\alpha \right)}{\mathrm{g}}$$$$\mathrm{where}\mathrm{e}=\mathrm{coefficient}\mathrm{of}\mathrm{restitution}.$$

Commented by Tawa11 last updated on 14/Sep/21

$$\mathrm{nice}$$

Answered by mr W last updated on 04/Apr/21

Commented by mr W last updated on 04/Apr/21

$$u_n\cos {\theta }_n=u_{n-1}\cos {\theta }_{n-1}=u\cos \alpha$$$$u_n\sin {\theta }_n=e{u}_{n-1}\sin {\theta }_{n-1}$$$$$$$$\tan {\theta }_n=e\tan {\theta }_{n-1}=e^2\tan {\theta }_{n-2}=...$$$$=e^{n-1}\tan {\theta }_1=e^{n-1}\tan \alpha$$$$R_n=2u_n\cos {\theta }_n\times \frac{u_n\sin {\theta }_n}{g}$$$$=\frac{2{\left(u_n\cos {\theta }_n\right)}^2\tan {\theta }_n}{g}$$$$=\frac{2u^2{\cos }^2\alpha e^{n-1}\tan \alpha }{g}$$$$=\frac{2u^2\cos \alpha e^{n-1}\sin \alpha }{g}$$$$=\frac{e^{n-1}{u}^2\sin 2\alpha }{g}$$$$=e^{n-1}{R}_1$$$$R=\underset{k=1}{\sum ^n}{R}_k=(1+e+e^2+...e^{n-1})R_1=\frac{1-e^n}{1-e}{R}_1$$$$R_{max}=\underset{n\to \mathrm{\infty }}{\lim }\underset{k=1}{\sum ^n}{R}_k=\frac{R_1}{1-e}=\frac{u^2\sin 2\alpha }{(1-e)g}$$

Commented by I want to learn more last updated on 04/Apr/21

$$\mathrm{Wow},\mathrm{i}\mathrm{really}\mathrm{appreciate}\mathrm{sir}.\mathrm{God}\mathrm{bless}\mathrm{you}.$$

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