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Question Number 114512 by Rio Michael last updated on 19/Sep/20

Commented by Rio Michael last updated on 19/Sep/20

$$\mathrm{need}\mathrm{some}\mathrm{help}\mathrm{with}\mathrm{the}\mathrm{above},\mathrm{am}$$$$\mathrm{particularly}\mathrm{stuck}\mathrm{at}\left(\mathrm{e}\right)$$

Commented by Dwaipayan Shikari last updated on 19/Sep/20

$$R=\frac{v_0^{2}sin2\theta }{g}$$$$gR=v_0^{2}sin2\theta$$$$\theta =\frac{1}{2}{sin}^{-1}\frac{gR}{v_0^2}\Rightarrow \theta =\frac{1}{2}{sin}^{-1}\left(3{sin}^2\theta \right)$$$$\frac{{\left(v_{0}sin\theta \right)}^2}{2g}=\frac{R}{6}$$$$v_0^2{sin}^2\theta =\frac{Rg}{3}$$$$So$$$$sin2\theta =3{sin}^2\theta$$$$2sin\theta cos\theta -3{sin}^2\theta =0$$$$sin\theta =0\Rightarrow \theta =k\pi$$$$or$$$$2cos\theta =3sin\theta$$$$tan\theta =\frac{2}{3}\Rightarrow \theta =k\pi +{tan}^{-1}\frac{2}{3}$$

Commented by Rio Michael last updated on 19/Sep/20

$$\mathrm{God}\mathrm{bless}\mathrm{you}$$

Commented by Rio Michael last updated on 19/Sep/20

$$\mathrm{but}\mathrm{sir}\mathrm{what}\mathrm{if}{v}_0^2\mathrm{is}\mathrm{also}\mathrm{in}\mathrm{terms}\mathrm{of}{\theta }_0$$

Answered by mr W last updated on 19/Sep/20

$$Range=R$$$$max.heighth=\frac{R}{6}$$$$\Rightarrow \tan {\theta }_0=\frac{4h}{R}=\frac{2}{3}$$$$2gh=v_0^2{\sin }^2{\theta }_0=v_0^2{\left(\frac{2}{\sqrt{13}}\right)}^2$$$$2g\frac{R}{6}=\frac{4}{13}{v}_0^2$$$$\Rightarrow v_0=\frac{1}{2}\sqrt{\frac{13gR}{3}}$$$$\left(a\right)$$$$t=\frac{R}{v_0\cos {\theta }_0}=\frac{R}{\frac{1}{2}\sqrt{\frac{13gR}{3}}\times \frac{3}{\sqrt{13}}}=2\sqrt{\frac{R}{3g}}$$$$\left(b\right)$$$$atpeak{v}_y=0$$$$v=v_x=v_0\cos {\theta }_0=\frac{1}{2}\sqrt{3gR}$$$$\left(c\right)$$$$v_{0,y}=v_0\sin {\theta }_0=\frac{1}{2}\sqrt{2gR}$$$$\left(d\right)$$$$v_0=\frac{1}{2}\sqrt{\frac{13gR}{3}}$$$$\left(e\right)$$$${\theta }_0={\tan }^{-1}\frac{2}{3}\approx 33.7°$$$$\left(f\right)$$$$at{\theta }_0=90°:$$$$h_{max}=\frac{v_0^2}{2g}=\frac{13R}{24}$$$$\left(g\right)$$$$at{\theta }_0=45°$$$$R_{max}=\frac{v_0^2}{g}=\frac{13R}{12}$$$$\left(h\right)$$$$d=2v_0\cos {\theta }_{0}t$$

Commented by Rio Michael last updated on 20/Sep/20

$$\mathrm{thank}\mathrm{you}\mathrm{Mr}\mathcal{W},\mathrm{but}\mathrm{i}\mathrm{must}\mathrm{ask}$$$$\mathrm{why}\mathrm{did}\mathrm{you}\mathrm{use}{\theta }_0\mathrm{just}\mathrm{for}\mathrm{the}$$$$\mathrm{first}\mathrm{quadrant}\mathrm{and}{\mathrm{didn}}^{\prime }\mathrm{t}\mathrm{give}\mathrm{a}\mathrm{general}$$$$\mathrm{formular}\mathrm{like}{\theta }_0=\{\begin{array}{l}\pi n\\ \pi n+{\tan }^{-1}\left(\frac{2}{3}\right)\end{array}n\in {\mathbb{Z}}^{\ast }$$$$\mathrm{is}\mathrm{there}\mathrm{some}\mathrm{reason}\mathrm{or}\mathrm{explanation}?$$

Commented by mr W last updated on 21/Sep/20

$$hereisphysics,notmathematics!$$

Commented by mr W last updated on 21/Sep/20

$$i{didn}^{\prime }texpectthatyouaskthis.$$$$ididexpectthatyouaskhowigot$$$$directly\tan \theta =\frac{2}{3}.$$

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