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Question Number 18077 by virus last updated on 15/Jul/17

$$Amangoinatreeislocated(30,40)fromthe$$$$pointofprojectionofstone.Findthe$$$$minimumspeedandtheangleofprojevtion$$$$ofthestonesoastohitthemango$$

Answered by ajfour last updated on 15/Jul/17

$$\mathrm{Equation}\mathrm{of}\mathrm{trajectory}\mathrm{is}$$$$\mathrm{y}=\mathrm{xtan}\theta -\frac{{\mathrm{gx}}^2}{{2\mathrm{u}}^2}(1+\tan {}^2\theta )$$$$\mathrm{point}\mathrm{M}(30,40)\mathrm{must}\mathrm{lie}\mathrm{on}\mathrm{it};\mathrm{so}$$$$40=30\tan \theta -\frac{10\left(900\right)}{{2\mathrm{u}}^2}(1+\tan {}^2\theta )$$$$\mathrm{differentiating}\mathrm{and}\mathrm{equating}\mathrm{to}\mathrm{zero}:$$$$0=30\sec {}^2\theta -\frac{9000}{{2\mathrm{u}}^2}\left(2\tan \theta \sec {}^2\theta \right)$$$$\Rightarrow \tan \theta =\frac{{30\mathrm{u}}^2}{9000}=\frac{{\mathrm{u}}^2}{300}$$$$\Rightarrow 40=30\left(\frac{{\mathrm{u}}^2}{300}\right)-\frac{9000}{{2\mathrm{u}}^2}(1+\frac{{\mathrm{u}}^4}{90000})$$$$\Rightarrow 40=\frac{{\mathrm{u}}^2}{10}-\frac{9000}{{2\mathrm{u}}^2}-\frac{{\mathrm{u}}^2}{20}$$$$40=\frac{{\mathrm{u}}^2}{20}-\frac{9000}{{2\mathrm{u}}^2}(..\times {20\mathrm{u}}^2\mathrm{gives})$$$${800\mathrm{u}}^2={\mathrm{u}}^4-90000$$$${({\mathrm{u}}^2-400)}^2=90000+160000$$$${\mathrm{u}}^2-400=\sqrt{25\times {10000}_{}}$$$${\mathrm{u}}^2=900\Rightarrow {\mathrm{u}}_{\min }=30\mathrm{m}/\mathrm{s},$$$$\theta ={\tan }^{-1}\left(\frac{{\mathrm{u}}^2}{300}\right)={\tan }^{-1}3.$$

Commented by virus last updated on 15/Jul/17

$$thankyousir$$

Commented by ajfour last updated on 15/Jul/17

$$\mathrm{Is}\mathrm{answer}\mathrm{correct},\mathrm{virus}?$$

Commented by virus last updated on 15/Jul/17

$$yessir$$

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