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Question Number 18261 by Tinkutara last updated on 17/Jul/17

$$\mathrm{What}\mathrm{is}\mathrm{the}\mathrm{maximum}\mathrm{angle}\mathrm{to}\mathrm{the}$$$$\mathrm{horizontal}\mathrm{at}\mathrm{which}\mathrm{a}\mathrm{stone}\mathrm{can}\mathrm{be}$$$$\mathrm{thrown}\mathrm{and}\mathrm{always}\mathrm{be}\mathrm{moving}\mathrm{away}$$$$\mathrm{from}\mathrm{the}\mathrm{thrower}?$$

Answered by ajfour last updated on 26/Jul/17

$$\mathrm{r}=\sqrt{{\mathrm{x}}^2+{\mathrm{y}}^2},\mathrm{x}={\mathrm{u}}_{\mathrm{x}}\mathrm{t},\mathrm{y}={\mathrm{u}}_{\mathrm{y}}\mathrm{t}-\frac{{\mathrm{gt}}^2}{2}$$$$\frac{\mathrm{dr}}{\mathrm{dt}}=\frac{(\frac{2\mathrm{xdx}}{\mathrm{dt}}+\frac{2\mathrm{ydy}}{\mathrm{dt}})}{2\sqrt{{\mathrm{x}}^2+{\mathrm{y}}^2}}$$$$=\frac{{\mathrm{u}}_{\mathrm{x}}^2\mathrm{t}+({\mathrm{u}}_{\mathrm{y}}\mathrm{t}-\frac{{\mathrm{gt}}^2}{2})({\mathrm{u}}_{\mathrm{y}}-\mathrm{gt})}{1\sqrt{{\mathrm{x}}^2+{\mathrm{y}}^2}}$$$$\tan \theta =\frac{{\mathrm{u}}_{\mathrm{y}}}{{\mathrm{u}}_{\mathrm{x}}}$$$$\frac{\mathrm{dr}}{\mathrm{dt}}>0\Rightarrow \mathrm{t}[{\mathrm{u}}_{\mathrm{x}}^2+({\mathrm{u}}_{\mathrm{y}}-\frac{\mathrm{gt}}{2})({\mathrm{u}}_{\mathrm{y}}-\mathrm{gt})]>0$$$$\mathrm{or}\frac{{\mathrm{g}}^2{\mathrm{t}}^2}{2}-\frac{{3\mathrm{u}}_{\mathrm{y}}\mathrm{gt}}{2}+{\mathrm{u}}_{\mathrm{x}}^2+{\mathrm{u}}_{\mathrm{y}}^2>0$$$$\Rightarrow \mathrm{D}=\frac{{9\mathrm{u}}_{\mathrm{y}}^2{\mathrm{g}}^2}{4}-{2\mathrm{g}}^2({\mathrm{u}}_{\mathrm{x}}^2+{\mathrm{u}}_{\mathrm{y}}^2)<0$$$$\Rightarrow \frac{{\mathrm{u}}_{\mathrm{y}}}{\sqrt{{\mathrm{u}}_{\mathrm{x}}^2+{\mathrm{u}}_{\mathrm{y}}^2}}<\frac{2\sqrt{2}}{3}$$$$\mathrm{or}\frac{{\mathrm{u}}_{\mathrm{y}}}{\mathrm{u}}<\frac{2\sqrt{2}}{3}\Rightarrow \sin \theta <\frac{2\sqrt{2}}{3}$$$$\tan \theta <2\sqrt{2}\cos \theta >\frac{1}{3}$$$$\theta <{\cos }^{-1}\left(1/3\right)\mathrm{which}\mathrm{is}\mathrm{same}\mathrm{as}$$$$\theta <{\tan }^{-1}\left(2\sqrt{2}\right).$$

Commented by Tinkutara last updated on 26/Jul/17

$$\mathrm{Thanks}\mathrm{Sir}!$$

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