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Question Number 213234 by ajfour last updated on 01/Nov/24

Commented by Ghisom last updated on 01/Nov/24

$$\mathrm{let}r=1$$$$P\in \mathrm{circle}:P=\left(\begin{array}{c}\cos \theta \\ 1+\sin \theta \end{array}\right)$$$$\mathrm{parabola}:y=(\frac{2+\sin \theta }{\cos \theta }-\frac{x}{{\cos }^2\theta })x$$$$\tan \alpha =\frac{2+\sin \theta }{\cos \theta }$$$$$$$$\mathrm{with}\theta =\arctan \frac{3}{4}\mathrm{we}\mathrm{get}$$$$P=\left(\begin{array}{c}4/5\\ 8/5\end{array}\right)$$$$\mathrm{par}:y=(\frac{13}{4}-\frac{25}{16}x)x$$$$\tan \alpha =\frac{13}{4}$$

Commented by mr W last updated on 02/Nov/24

$$pleasepostyouransweras‘‘{answer}^{″},$$$$notas‘‘{comment}^{″}.$$

Answered by a.lgnaoui last updated on 01/Nov/24

$$\mathbf{C}\mathrm{alcul}\mathrm{de}\tan \mathit{\alpha }$$$$\mathrm{Mouvfment}\mathrm{de}\mathbf{A}\to \mathbf{C}$$$$\mathrm{a}=\mathrm{g}={\mathrm{g}}_{\mathrm{y}}=+\mathrm{g}\Rightarrow {\mathrm{v}}_{\mathrm{y}}=\mathrm{gt}+{\mathrm{v}}_0$$$${\mathrm{a}}_{\mathrm{x}}=0\Rightarrow {\mathrm{v}}_{\mathrm{x}}=\mathrm{cte}=\frac{4}{5}\mathrm{R}=\mathrm{Rcos}\theta$$$$\tan \theta =\frac{3}{4}\Rightarrow \theta ={\tan }^{-1}\left(\frac{3}{4}\right)\{\begin{array}{l}\cos \theta =\frac{4}{5}\\ \sin \theta =\frac{3}{5}\end{array}$$$${\mathrm{v}}_0=\{\begin{array}{l}{\mathrm{v}}_{0\mathrm{x}}=\frac{4\mathrm{R}}{5}\\ {\mathrm{v}}_{0\mathrm{y}}=\frac{3\mathrm{R}}{5}\end{array}$$$$\mid \mid {\mathrm{V}}_0\mid \mid =\sqrt{{\mathrm{v}}_{0\mathrm{x}}^2+{\mathrm{v}}_{0\mathrm{y}}^2}=\mathrm{R}$$$$\mathrm{A}(0,0)\mathrm{B}(\frac{4\mathrm{R}}{5},0);(\mathrm{C}(\frac{4\mathrm{R}}{5};\frac{8\mathrm{R}}{5})$$$$\{\begin{array}{l}{\mathrm{v}}_{\mathrm{y}}=\mathrm{gt}+\frac{3\mathrm{R}}{5}\mathrm{y}=\frac{1}{2}{\mathrm{gt}}^2+\frac{3\mathrm{R}}{5}\mathrm{t}+{\mathrm{y}}_0\\ {\mathrm{v}}_{\mathrm{x}}=\frac{4\mathrm{R}}{5}={\mathrm{v}}_{0\mathrm{x}}\mathrm{x}=\frac{4\mathrm{R}}{5}\mathrm{t}\end{array}$$$$\mathrm{au}\mathrm{point}\mathrm{C}(\mathrm{x}=\frac{4\mathrm{R}}{5};\mathrm{y}=\frac{8\mathrm{R}}{5})$$$$\frac{8\mathrm{R}}{5}=\frac{1}{2}{\mathrm{gt}}^2+\frac{3\mathrm{R}}{5}\mathrm{t}\mathrm{x}=\frac{4\mathrm{R}}{5}\mathrm{t}$$$${\mathrm{v}}^2-{\mathrm{v}}_0^2=2\mathrm{g}(\mathrm{x}-{\mathrm{x}}_0)=2\mathrm{gx}$$$$\mid \mid {\mathrm{v}}^2\mid \mid ={\mathrm{v}}_{\mathrm{x}}^2+{\mathrm{v}}_{\mathrm{y}}^2=\frac{{16\mathrm{R}}^2}{25}+\frac{8\mathbf{g}\mathrm{R}}{5}=\frac{8\mathrm{R}}{5}(\frac{2\mathrm{R}}{5}+\mathrm{g})$$$$\tan \alpha =\frac{{\mathrm{v}}_{\mathrm{y}}}{{\mathrm{v}}_{\mathrm{x}}}=\frac{4\mathrm{R}}{5}+2\mathbf{g}$$$$(forR=5\mathrm{and}\mathrm{g}=10\alpha =87.61\tan \alpha =24)$$$$$$

Commented by a.lgnaoui last updated on 01/Nov/24

Commented by mr W last updated on 01/Nov/24

$$verywrong!$$$$\tan \alpha hasnounit.$$$$Rhastheunitoflength\left[m\right].$$$$ghastheunitofacceleration\left[m/s^2\right].$$$$\tan \alpha =\frac{4R}{5}+2gmakesneversense!$$$$justlikewecannotsay$$$$theangleisequaltothesumfrom$$$$4metersand2kilograms.$$

Answered by mr W last updated on 01/Nov/24

Commented by mr W last updated on 01/Nov/24

$$suchthattheballreturnsinthe$$$$samepath,theballmusthitthe$$$$wallperpendicularly.$$$$x=u\cos \alpha t$$$$y=u\sin \alpha t-\frac{{gt}^2}{2}$$$$\Rightarrow y=\tan \alpha x-\frac{g(1+{\tan }^2\alpha )x^2}{2u^2}$$$$letm=\tan \alpha$$$$y=mx-\frac{g(1+m^2)x^2}{2u^2}$$$$\frac{dy}{dx}=m-\frac{g(1+m^2)x}{u^2}$$$$\frac{8R}{5}=m\times \frac{4R}{5}-\frac{g(1+m^2)}{2u^2}\times \frac{16R^2}{25}$$$$\Rightarrow 2=m-\frac{2gR(1+m^2)}{5u^2}...\left(i\right)$$$$\tan \phi =m-\frac{g(1+m^2)}{u^2}\times \frac{4R}{5}$$$$\Rightarrow \frac{3}{4}=m-\frac{4gR(1+m^2)}{5u^2}...\left(iii\right)$$$$\left(i\right)\times 2-\left(ii\right):$$$$4-\frac{3}{4}=m$$$$\Rightarrow \tan \alpha =\frac{13}{4}\Rightarrow \alpha ={\tan }^{-1}\frac{13}{4}\approx 72.9°$$

Commented by a.lgnaoui last updated on 01/Nov/24

$$\mathrm{good}\mathrm{thank}\mathrm{you}$$$$\mathrm{I}\mathrm{ask}\mathrm{If}{\mathrm{v}}_{\mathrm{y}}={\mathrm{v}}_0(\alpha =60+\phi )?$$

Commented by mr W last updated on 01/Nov/24

$$no.$$

Commented by ajfour last updated on 01/Nov/24

$$sayithitswithspeedv.$$$$v\cos \phi =u\cos \alpha$$$$v^2=\frac{u^2\cos {}^2\alpha }{\cos {}^2\phi }=u^2-2gR(1+\sin \phi )$$$$\Rightarrow \frac{2gR\cos {}^2\phi }{u^2\cos {}^2\alpha }=(1-\frac{\cos {}^2\phi }{\cos {}^2\alpha })\left(\frac{1}{1+\sin \phi }\right)$$$$R(1+\sin \phi )$$$$=R\cos \phi \tan \alpha -\frac{{gR}^2\cos {}^2\phi }{2u^2\cos {}^2\alpha }$$$$1+\sin \phi =\cos \phi \tan \alpha$$$$+\frac{1}{4}(1-\frac{\cos {}^2\phi }{\cos {}^2\alpha })\frac{1}{(1+\sin \phi )}$$$$say\tan \alpha =m$$$$\frac{8}{5}=\frac{4m}{5}+\frac{1}{4}\times \frac{5}{8}-\frac{4}{25}\times \frac{5}{8}(1+m^2)$$$$m^2+1=8m-16+\frac{25}{16}$$$${(m-4)}^2=\frac{9}{16}$$$$m=4\pm \frac{3}{4}$$$$\Rightarrow \mathbf{tan}\mathit{\alpha }=4.75or3.25$$$$★$$

Commented by ajfour last updated on 01/Nov/24

$$ThankyoumrWsir,Igottwo$$$$answers,bothihopearecorrect;$$$$oneissameasyours.$$

Commented by Ghisom last updated on 01/Nov/24

$$\tan \alpha =4.75\mathrm{is}\mathrm{wrong}$$$$\mathrm{you}\mathrm{squared}\mathrm{in}\mathrm{line}2(v^2=...)\mathrm{thus}\mathrm{introducing}$$$$\mathrm{this}\mathrm{false}\mathrm{solution}$$

Commented by mr W last updated on 01/Nov/24

$$onlywithm=\frac{13}{4}theballreturnsback$$$$followingthesamepath,thegreen$$$$path.$$

Commented by mr W last updated on 01/Nov/24

Commented by ajfour last updated on 02/Nov/24

$$yes,thankyou;soitindeedseems.$$

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