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Question Number 154547 by peter frank last updated on 19/Sep/21

$$\mathrm{A}\mathrm{particle}\mathrm{is}\mathrm{projected}\mathrm{inside}\mathrm{the}\mathrm{tunnel}$$$$\mathrm{which}\mathrm{is}4\mathrm{m}\mathrm{high}.\mathrm{if}\mathrm{the}\mathrm{initial}\mathrm{speed}$$$$\mathrm{is}{\mathrm{V}}_{\mathrm{o}}.\mathrm{show}\mathrm{that}\mathrm{the}\mathrm{maximum}$$$$\mathrm{range}\mathrm{inside}\mathrm{the}\mathrm{tunnel}\mathrm{is}\mathrm{given}$$$$\mathrm{by}\mathrm{R}=4\sqrt{2}\sqrt{\frac{{\mathrm{V}}_{\mathrm{o}}^2}{\mathrm{g}}-8}$$

Answered by peter frank last updated on 20/Sep/21

Commented by peter frank last updated on 20/Sep/21

$$\mathrm{Range}=\frac{{2\mathrm{v}}_{\mathrm{o}}^2\sin \theta \cos \theta }{\mathrm{g}}$$$${\mathrm{H}}_{\max }=\frac{{\mathrm{v}}_{\mathrm{o}}^2\sin {}^2\theta }{\mathrm{g}}=4\mathrm{m}$$$$\sin {}^2\theta =\frac{8\mathrm{g}}{{\mathrm{v}}_{\mathrm{o}}^2}$$$$\sin \theta =\frac{2\sqrt{2\mathrm{g}}}{{\mathrm{v}}_{\mathrm{o}}}$$$$\cos {}^2\theta =\sqrt{1-\sin {}^2\theta }$$$$\cos \theta =\frac{1}{{\mathrm{v}}_{\mathrm{o}}}\sqrt{{\mathrm{v}}_{\mathrm{o}}^2-8\mathrm{g}}$$$$\mathrm{Range}=\frac{{2\mathrm{v}}_{\mathrm{o}}^2\sin \theta \cos \theta }{\mathrm{g}}$$$$\mathrm{Range}=\frac{{2\mathrm{v}}_{\mathrm{o}}^2\frac{2\sqrt{2\mathrm{g}}}{{\mathrm{v}}_{\mathrm{o}}}\frac{1}{{\mathrm{v}}_{\mathrm{o}}}\sqrt{{\mathrm{v}}_{\mathrm{o}}^2-8\mathrm{g}}}{\mathrm{g}}$$$$\mathrm{R}=\frac{4\sqrt{2\mathrm{g}}}{\mathrm{g}}\sqrt{{\mathrm{v}}_{\mathrm{o}}^2-8\mathrm{g}}$$$$\mathrm{R}=4\sqrt{2}\frac{\sqrt{\mathrm{g}}}{\mathrm{g}}\sqrt{{\mathrm{v}}_{\mathrm{o}}^2-8\mathrm{g}}$$$$\mathrm{but}\frac{\sqrt{\mathrm{g}}}{\mathrm{g}}=\sqrt{\frac{\mathrm{g}}{{\mathrm{g}}^2}}$$$$\mathrm{R}=4\sqrt{2}.\sqrt{\frac{\mathrm{g}}{{\mathrm{g}}^2}{\mathrm{v}}_{\mathrm{o}}^2-8\mathrm{g}}$$$$\mathrm{R}=4\sqrt{2}.\sqrt{\frac{1}{\mathrm{g}}.{\mathrm{v}}_{\mathrm{o}}^2-8\mathrm{g}}$$$$\mathrm{R}=4\sqrt{2}.\sqrt{\frac{{\mathrm{v}}_{\mathrm{o}}^2}{\mathrm{g}}-8}$$$$$$

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

$$wellsolved!$$

Commented by peter frank last updated on 21/Sep/21

$$\mathrm{thank}\mathrm{you}.$$

Commented by Tawa11 last updated on 21/Sep/21

$$\mathrm{nice}\mathrm{sir}$$

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