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Question Number 133487 by mr W last updated on 22/Feb/21

Commented by mr W last updated on 22/Feb/21

$$aballisprojectedalongthesmooth$$$$floorandreturnsbackaftertwo$$$$timesimpactwiththecircularwall.$$$$iftherestitutioncoefficientise,$$$$findtheanglesof\mathrm{\Delta }ABC.$$

Answered by mr W last updated on 22/Feb/21

Commented by mr W last updated on 23/Feb/21

$$atpointB:$$$$\tan \beta =\frac{\tan \alpha }{e}$$$$atpointC:$$$$\tan \gamma =\frac{\tan \beta }{e}=\frac{\tan \alpha }{e^2}$$$$2(\alpha +\beta +\gamma =\pi$$$$\Rightarrow \alpha +\beta =\frac{\pi }{2}-\gamma$$$$\tan (\alpha +\beta )=\tan (\frac{\pi }{2}-\gamma )=\frac{1}{\tan \gamma }$$$$\frac{\tan \alpha +\frac{\tan \alpha }{e}}{1-\frac{{\tan }^2\alpha }{e}}=\frac{1}{\frac{\tan \alpha }{e^2}}$$$$\frac{(1+e)\tan \alpha }{e-{\tan }^2\alpha }=\frac{e^2}{\tan \alpha }$$$$(1+e+e^2){\tan }^2\alpha =e^3$$$$\Rightarrow \tan \alpha =e\sqrt{\frac{e}{1+e+e^2}}$$$$\Rightarrow \tan \beta =\sqrt{\frac{e}{1+e+e^2}}$$$$\Rightarrow \tan \gamma =\frac{1}{e}\sqrt{\frac{e}{1+e+e^2}}$$$$\tan (\alpha +\beta )=\frac{(1+e)\sqrt{\frac{e}{1+e+e^2}}}{1-e\times \frac{e}{1+e+e^2}}=\sqrt{e(1+e+e^2)}$$$$\tan (\beta +\gamma )=\frac{(1+\frac{1}{e})\sqrt{\frac{e}{1+e+e^2}}}{1-\frac{1}{e}\times \frac{e}{1+e+e^2}}=\frac{\sqrt{e(1+e+e^2)}}{e^2}$$$$\tan (\gamma +\alpha )=\frac{(\frac{1}{e}+e)\sqrt{\frac{e}{1+e+e^2}}}{1-\frac{e}{1+e+e^2}}=\frac{\sqrt{e(1+e+e^2)}}{e}$$$$\Rightarrow {\theta }_1=\gamma +\alpha ={\tan }^{-1}\frac{\sqrt{e(1+e+e^2)}}{e}$$$$\Rightarrow {\theta }_2=\alpha +\beta ={\tan }^{-1}\sqrt{e(1+e+e^2)}$$$$\Rightarrow {\theta }_3=\beta +\gamma ={\tan }^{-1}\frac{\sqrt{e(1+e+e^2)}}{e^2}$$$$example:e=0.5$$$${\theta }_1={\tan }^{-1}\frac{\sqrt{14}}{2}=61.87°$$$${\theta }_2={\tan }^{-1}\frac{\sqrt{14}}{4}=43.09°$$$${\theta }_3={\tan }^{-1}\sqrt{14}=75.04°$$

Commented by mr W last updated on 23/Feb/21

Commented by mr W last updated on 23/Feb/21

Commented by mr W last updated on 23/Feb/21

Commented by mr W last updated on 23/Feb/21

Commented by mr W last updated on 23/Feb/21

$$\mathit{b}\mathit{a}\mathit{c}\mathit{k}\mathit{g}\mathit{r}\mathit{o}\mathit{u}\mathit{n}\mathit{d}\mathit{k}\mathit{n}\mathit{o}\mathit{w}\mathit{l}\mathit{e}\mathit{d}\mathit{g}\mathit{e}:$$$$\left(seealsoQ132102\right)$$

Commented by mr W last updated on 23/Feb/21

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