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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

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

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