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Question Number 155292 by peter frank last updated on 28/Sep/21

Answered by peter frank last updated on 30/Sep/21

R=a=((v^2 sin 2θ)/g)=((2v^2 sin θcos θ)/g)...(i) H=x=((v^2 sin^2 θ)/(2g)) cosec^2 θ=(v^2 /(2gx))..(ii) (x/a)=(((2v^2 sin θcos θ)/g)/((v^2 sin^2 θ)/(2g)))=4cot θ cot θ=(a/(4x)) cosec^2 θ=1+cot^2 θ (v^2 /(2gx))=1+((a/(4x)))^2 v^2 =2gx+((ga^2 )/(16x^2 )).2gx v^2 =2gx+((ga^2 )/(8x)) 16x^2 g−8v^2 x+ga^2 =0

$$\mathrm{R}=\mathrm{a}=\frac{\mathrm{v}^{\mathrm{2}} \mathrm{sin}\:\mathrm{2}\theta}{\mathrm{g}}=\frac{\mathrm{2v}^{\mathrm{2}} \mathrm{sin}\:\theta\mathrm{cos}\:\theta}{\mathrm{g}}...\left(\mathrm{i}\right) \\ $$$$\mathrm{H}=\mathrm{x}=\frac{\mathrm{v}^{\mathrm{2}} \mathrm{sin}\:^{\mathrm{2}} \theta}{\mathrm{2g}} \\ $$$$\mathrm{cosec}\:^{\mathrm{2}} \theta=\frac{\mathrm{v}^{\mathrm{2}} }{\mathrm{2gx}}..\left(\mathrm{ii}\right) \\ $$$$\frac{\mathrm{x}}{\mathrm{a}}=\frac{\frac{\mathrm{2v}^{\mathrm{2}} \mathrm{sin}\:\theta\mathrm{cos}\:\theta}{\mathrm{g}}}{\frac{\mathrm{v}^{\mathrm{2}} \mathrm{sin}\:^{\mathrm{2}} \theta}{\mathrm{2g}}}=\mathrm{4cot}\:\theta \\ $$$$\mathrm{cot}\:\theta=\frac{\mathrm{a}}{\mathrm{4x}} \\ $$$$\mathrm{cosec}\:^{\mathrm{2}} \theta=\mathrm{1}+\mathrm{cot}\:^{\mathrm{2}} \theta \\ $$$$\frac{\mathrm{v}^{\mathrm{2}} }{\mathrm{2gx}}=\mathrm{1}+\left(\frac{\mathrm{a}}{\mathrm{4x}}\right)^{\mathrm{2}} \\ $$$$\mathrm{v}^{\mathrm{2}} =\mathrm{2gx}+\frac{\mathrm{ga}^{\mathrm{2}} }{\mathrm{16x}^{\mathrm{2}} }.\mathrm{2gx} \\ $$$$\mathrm{v}^{\mathrm{2}} =\mathrm{2gx}+\frac{\mathrm{ga}^{\mathrm{2}} }{\mathrm{8x}} \\ $$$$\mathrm{16x}^{\mathrm{2}} \mathrm{g}−\mathrm{8v}^{\mathrm{2}} \mathrm{x}+\mathrm{ga}^{\mathrm{2}} =\mathrm{0} \\ $$$$ \\ $$

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