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Question Number 16455 by Tinkutara last updated on 22/Jun/17

$$\mathrm{The}\mathrm{speed}\mathrm{of}\mathrm{a}\mathrm{projectile}\mathrm{when}\mathrm{it}\mathrm{is}\mathrm{at}$$$$\mathrm{its}\mathrm{greatest}\mathrm{height}\mathrm{is}\sqrt{\frac{2}{5}}\mathrm{times}\mathrm{its}$$$$\mathrm{speed}\mathrm{at}\mathrm{half}\mathrm{the}\mathrm{maximum}\mathrm{height}.$$$$\mathrm{What}\mathrm{is}\mathrm{its}\mathrm{angle}\mathrm{of}\mathrm{projection}?$$

Answered by sandy_suhendra last updated on 23/Jun/17

$$\mathrm{A}=\mathrm{point}\mathrm{of}\mathrm{the}\mathrm{maximum}\mathrm{height}$$$$\mathrm{B}=\mathrm{point}\mathrm{of}\mathrm{half}\mathrm{maximum}\mathrm{height}$$$${\mathrm{V}}_{\mathrm{A}}={\mathrm{V}}_0\cos \alpha$$$${\mathrm{h}}_{\mathrm{A}}=\frac{{\mathrm{V}}_0^2{\sin }^2\alpha }{2\mathrm{g}}$$$${\mathrm{V}}_{\mathrm{A}}=\sqrt{\frac{2}{5}}{\mathrm{V}}_{\mathrm{B}}\Rightarrow {\mathrm{V}}_{\mathrm{B}}=\frac{5}{2}\sqrt{\frac{2}{5}}{\mathrm{V}}_{\mathrm{A}}=\frac{5}{2}\sqrt{\frac{2}{5}}{\mathrm{V}}_0\cos \alpha$$$${\mathrm{h}}_{\mathrm{B}}=\frac{1}{2}{\mathrm{h}}_{\mathrm{A}}=\frac{{\mathrm{V}}_0^2{\sin }^2\alpha }{4\mathrm{g}}$$$${\mathrm{KE}}_{\mathrm{A}}+{\mathrm{PE}}_{\mathrm{A}}={\mathrm{KE}}_{\mathrm{B}}+{\mathrm{PE}}_{\mathrm{B}}$$$$\frac{1}{2}\mathrm{m}.{\mathrm{V}}_{\mathrm{A}}^2+\mathrm{m}.\mathrm{g}.{\mathrm{h}}_{\mathrm{A}}=\frac{1}{2}\mathrm{m}.{\mathrm{V}}_{\mathrm{B}}^2+\mathrm{m}.\mathrm{g}.{\mathrm{h}}_{\mathrm{B}}$$$$\frac{1}{2}\mathrm{m}{\left({\mathrm{V}}_0\cos \alpha \right)}^2+\mathrm{m}.\mathrm{g}.\frac{{\mathrm{V}}_0^2{\sin }^2\alpha }{2\mathrm{g}}=\frac{1}{2}\mathrm{m}{\left(\frac{5}{2}\sqrt{\frac{2}{5}}{\mathrm{V}}_0\cos \alpha \right)}^2+\mathrm{m}.\mathrm{g}.\frac{{\mathrm{V}}_0^2{\sin }^2\alpha }{4\mathrm{g}}$$$$\frac{1}{2}{\cos }^2\alpha +\frac{1}{2}{\sin }^2\alpha =\frac{5}{4}{\cos }^2\alpha +\frac{1}{4}{\sin }^2\alpha$$$$\frac{1}{4}{\sin }^2\alpha =\frac{3}{4}{\cos }^2\alpha$$$${\tan }^2\alpha =3\Rightarrow \tan \alpha =\sqrt{3}\Rightarrow \alpha =60°$$

Commented by Tinkutara last updated on 23/Jun/17

$$\mathrm{But}\mathrm{answer}\mathrm{is}60°.$$

Commented by sandy_suhendra last updated on 23/Jun/17

$$\mathrm{sorry},{\mathrm{I}}^{\prime }\mathrm{ve}\mathrm{a}\mathrm{mistake}\mathrm{at}\mathrm{the}\mathrm{second}\mathrm{line}\mathrm{from}\mathrm{the}\mathrm{bottom}$$$$\mathrm{but}{\mathrm{I}}^{\prime }\mathrm{ve}\mathrm{corrected}$$

Commented by Tinkutara last updated on 24/Jun/17

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

Answered by ajfour last updated on 23/Jun/17

$$Athalfthemax.height:$$$$\frac{1}{2}{mv}^2=\frac{1}{2}{mu}^2-\frac{mgH}{2}$$$$v^2=u^2-\frac{u^2\sin {}^2\theta }{2}...\left(i\right)$$$$Alsoasgivenu\cos \theta =\sqrt{\frac{2}{5}}v$$$$or{v}^2=\frac{5}{2}{u}^2\cos {}^2\theta \left(ii\right)$$$$equating\left(i\right)with\left(ii\right)$$$$u^2-\frac{u^2\sin {}^2\theta }{2}=\frac{5}{2}{u}^2\cos {}^2\theta$$$$(1+\tan {}^2\theta )-\frac{1}{2}\tan {}^2\theta =\frac{5}{2}$$$$\frac{1}{2}\tan {}^2\theta =\frac{3}{2}\Rightarrow \tan \theta =\pm \sqrt{3}$$$$\theta =60°,120°.$$$$$$

Commented by Tinkutara last updated on 24/Jun/17

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

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