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Question Number 21441 by Tinkutara last updated on 23/Sep/17
Find α in terms of θ using the equations: (i) u^2 sin^2 α = 2gd cos θ (ii) t = ((u cos α)/(g sin θ)) (iii) −d = ut sin α − ((gt^2 sin θ)/2)
$$\mathrm{Find}\:\alpha\:\mathrm{in}\:\mathrm{terms}\:\mathrm{of}\:\theta\:\mathrm{using}\:\mathrm{the}\:\mathrm{equations}: \\ $$$$\left({i}\right)\:{u}^{\mathrm{2}} \:\mathrm{sin}^{\mathrm{2}} \:\alpha\:=\:\mathrm{2}{gd}\:\mathrm{cos}\:\theta \\ $$$$\left({ii}\right)\:{t}\:=\:\frac{{u}\:\mathrm{cos}\:\alpha}{{g}\:\mathrm{sin}\:\theta} \\ $$$$\left({iii}\right)\:−{d}\:=\:{ut}\:\mathrm{sin}\:\alpha\:−\:\frac{{gt}^{\mathrm{2}} \:\mathrm{sin}\:\theta}{\mathrm{2}} \\ $$
Answered by $@ty@m last updated on 24/Sep/17
Substituting the value of t in (iii) −d = u(((u cos α)/(g sin θ)))sin α − g(((u cos α)/(g sin θ)))^2 ((sin)/2) −dgsin θ=u^2 cos αsin α−(1/2)u^2 cos^2 α −2dgsinθ=u^2 sin2α−u^2 cos^2 α −2dgsinθ=u^2 sin2α−u^2 (1−sin^2 α) u^2 −2dgsinθ=u^2 sin2α+u^2 sin^2 α u^2 −2dgsinθ=u^2 sin2α+2dgcosθ u^2 −2dgsinθ−2dgcosθ =u^2 sin2α u^2 sin2α=u^2 −2dg(sinθ+cosθ) sin2α=1−((2dg(sinθ+cosθ))/u^2 ) α=(1/2)sin^(−1) {1−((2dg(sinθ+cosθ))/u^2 ) }
$${Substituting}\:{the}\:{value}\:{of}\:{t}\:{in}\:\left({iii}\right) \\ $$$$−{d}\:=\:{u}\left(\frac{{u}\:\mathrm{cos}\:\alpha}{{g}\:\mathrm{sin}\:\theta}\right)\mathrm{sin}\:\alpha\:−\:{g}\left(\frac{{u}\:\mathrm{cos}\:\alpha}{{g}\:\mathrm{sin}\:\theta}\right)^{\mathrm{2}} \frac{\mathrm{sin}}{\mathrm{2}} \\ $$$$−{dg}\mathrm{sin}\:\theta={u}^{\mathrm{2}} \mathrm{cos}\:\alpha\mathrm{sin}\:\alpha−\frac{\mathrm{1}}{\mathrm{2}}{u}^{\mathrm{2}} \mathrm{cos}^{\mathrm{2}} \alpha \\ $$$$−\mathrm{2}{dg}\mathrm{sin}\theta={u}^{\mathrm{2}} \mathrm{sin2}\alpha−{u}^{\mathrm{2}} \mathrm{cos}^{\mathrm{2}} \alpha\:\:\: \\ $$$$−\mathrm{2}{dg}\mathrm{sin}\theta={u}^{\mathrm{2}} \mathrm{sin2}\alpha−{u}^{\mathrm{2}} \left(\mathrm{1}−\mathrm{sin}^{\mathrm{2}} \alpha\right)\: \\ $$$${u}^{\mathrm{2}} −\mathrm{2}{dg}\mathrm{sin}\theta={u}^{\mathrm{2}} \mathrm{sin2}\alpha+{u}^{\mathrm{2}} \mathrm{sin}^{\mathrm{2}} \alpha \\ $$$${u}^{\mathrm{2}} −\mathrm{2}{dg}\mathrm{sin}\theta={u}^{\mathrm{2}} \mathrm{sin2}\alpha+\mathrm{2}{dg}\mathrm{cos}\theta\: \\ $$$${u}^{\mathrm{2}} −\mathrm{2}{dg}\mathrm{sin}\theta−\mathrm{2}{dg}\mathrm{cos}\theta\:={u}^{\mathrm{2}} \mathrm{sin2}\alpha\: \\ $$$${u}^{\mathrm{2}} \mathrm{sin2}\alpha={u}^{\mathrm{2}} −\mathrm{2}{dg}\left(\mathrm{sin}\theta+\mathrm{cos}\theta\right)\:\:\: \\ $$$$\mathrm{sin2}\alpha=\mathrm{1}−\frac{\mathrm{2}{dg}\left(\mathrm{sin}\theta+\mathrm{cos}\theta\right)}{{u}^{\mathrm{2}} }\: \\ $$$$\alpha=\frac{\mathrm{1}}{\mathrm{2}}\mathrm{sin}^{−\mathrm{1}} \left\{\mathrm{1}−\frac{\mathrm{2}{dg}\left(\mathrm{sin}\theta+\mathrm{cos}\theta\right)}{{u}^{\mathrm{2}} }\:\right\} \\ $$
Commented by Tinkutara last updated on 24/Sep/17
What if the 3^(rd) equation was −d=utsin α−((gt^2 cos θ)/2)?
$$\mathrm{What}\:\mathrm{if}\:\mathrm{the}\:\mathrm{3}^{\mathrm{rd}} \:\mathrm{equation}\:\mathrm{was} \\ $$$$−{d}={ut}\mathrm{sin}\:\alpha−\frac{{gt}^{\mathrm{2}} \mathrm{cos}\:\theta}{\mathrm{2}}? \\ $$
Commented by $@ty@m last updated on 24/Sep/17
in that case t would be ((ucos α)/(gcos θ)) and the procedure would be similar.
$${in}\:{that}\:{case}\:{t}\:{would}\:{be}\:\frac{{u}\mathrm{cos}\:\alpha}{{g}\mathrm{cos}\:\theta} \\ $$$${and}\:{the}\:{procedure}\:{would}\:{be}\: \\ $$$${similar}. \\ $$
Commented by Tinkutara last updated on 24/Sep/17
Not everywhere cos θ, only in 3^(rd) equation.
$$\mathrm{Not}\:\mathrm{everywhere}\:\mathrm{cos}\:\theta,\:\mathrm{only}\:\mathrm{in}\:\mathrm{3}^{\mathrm{rd}} \\ $$$$\mathrm{equation}. \\ $$
Commented by $@ty@m last updated on 24/Sep/17
In that case it would result in an implicit function.
$${In}\:{that}\:{case}\:{it}\:{would}\:{result}\:{in}\: \\ $$$${an}\:{implicit}\:{function}. \\ $$
Commented by Tinkutara last updated on 24/Sep/17
Yes, can you solve it? Because it is required in a Physics question.
$$\mathrm{Yes},\:\mathrm{can}\:\mathrm{you}\:\mathrm{solve}\:\mathrm{it}?\:\mathrm{Because}\:\mathrm{it}\:\mathrm{is} \\ $$$$\mathrm{required}\:\mathrm{in}\:\mathrm{a}\:\mathrm{Physics}\:\mathrm{question}. \\ $$
Commented by $@ty@m last updated on 24/Sep/17
In an implicit fnction one variable cannot be expressed in terms of another variable. In my opnion, in such Physics question, there would be one more relation between the variables.
$${In}\:{an}\:{implicit}\:{fnction} \\ $$$${one}\:{variable}\:{cannot}\:{be}\:{expressed} \\ $$$${in}\:{terms}\:{of}\:{another}\:{variable}. \\ $$$${In}\:{my}\:{opnion}, \\ $$$${in}\:{such}\:{Physics}\:{question}, \\ $$$${there}\:{would}\:{be}\:{one}\:{more}\:{relation} \\ $$$${between}\:{the}\:{variables}. \\ $$

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