Question Number 72803 by mr W last updated on 03/Nov/19
Commented by mr W last updated on 03/Nov/19
$${a}\:{ball}\:{is}\:{thrown}\:{from}\:{point}\:{A}\:{with}\:{a} \\ $$$${speed}\:{u}.\:{after}\:{striking}\:{a}\:{point}\:{on}\:{the} \\ $$$${inclined}\:{plane},\:{it}\:{returns}\:{back}\:{to}\:{its} \\ $$$${original}\:{position}.\:{if}\:{the}\:{coefficient}\:{of} \\ $$$${restitution}\:{is}\:{e},\:{how}\:{large}\:{must}\:{u}\:{be} \\ $$$${such}\:{that}\:{the}\:{motion}\:{of}\:{the}\:{ball}\:{as} \\ $$$${described}\:{is}\:{possible}. \\ $$
Answered by ajfour last updated on 03/Nov/19
Commented by ajfour last updated on 04/Nov/19
$${It}\:{gets}\:{complicated}\:{Sir},\:{i}\:{tried} \\ $$$${but}\:{posted}\:{and}\:{deleted}.. \\ $$
Commented by mr W last updated on 04/Nov/19
$${thanks}\:{for}\:{trying}\:{sir}! \\ $$$${it}\:{is}\:{really}\:{complicated}. \\ $$
Commented by ajfour last updated on 04/Nov/19
![let tan θ=s, tan α=m, tan δ=n H=sD−((gD^2 (1+s^2 ))/(2u^2 )) ....(i) −H=nD−((gD^2 (1+n^2 ))/(2v_2 ^2 )) ...(ii) mD=h+H ...(iii) v_1 ^2 =u^2 −2gH ...(iv) v_2 cos φ=ev_1 cos ϕ v_2 sin φ=v_1 sin ϕ unknowns u, m, n, D, H, ϕ, φ v_2 ^2 =v_1 ^2 (e^2 cos^2 ϕ+sin^2 ϕ) ⇒ v_2 ^2 =(u^2 −2gH)(e^2 cos^2 ϕ+sin^2 ϕ) etan φ=tan ϕ tan δ=n=tan (90°−α−φ) ⇒ n=((1−tan φ)/(m+tan φ))=((e−tan ϕ)/(em+tan ϕ)) tan (90°−α+ϕ)=((√(u^2 sin^2 θ−2gH))/(ucos θ)) ⇒ ((1+mtan ϕ)/(m−tan ϕ))=(√(s^2 −((2gH(1+s^2 ))/u^2 ))) = f .....(v) 1+mtan ϕ=mf−ftan ϕ ⇒ tan ϕ=((mf−1)/(m+f)) ....(vi) n= ((e−((mf−1)/(m+f)))/(em+((mf−1)/(m+f))))=((em+ef−mf+1)/(em^2 +emf+mf−1)) ....(vii) from (ii) which is −H=nD−((gD^2 (1+n^2 ))/(2v_2 ^2 )) __________________________ −H=nD−((gD^2 (1+n^2 ){(mf−1)^2 +(m+f)^2 })/(2(u^2 −2gH)[e^2 +(mf−1)^2 ])) −H=(((em+ef−mf+1)/(em^2 +emf+mf−1)))D−((gD^2 {1+(((em+ef−mf+1)/(em^2 +emf+mf−1)))^2 }{(mf−1)^2 +(m+f)^2 })/(2(u^2 −2gH)[e^2 +(mf−1)^2 ])) ...(1) where f=(√(s^2 −((2gH(1+s^2 ))/u^2 ))) mD=h+H (2) ...(iii) H=sD−((gD^2 (1+s^2 ))/(2u^2 )) (3) ....(i) Now unknowns are u, s, H, D From (1), (2), (3) we obtain a correlation in u and s=tan θ ...........■](https://www.tinkutara.com/question/Q72897.png)
$${let}\:\mathrm{tan}\:\theta={s},\:\mathrm{tan}\:\alpha={m},\:\mathrm{tan}\:\delta={n} \\ $$$${H}={sD}−\frac{{gD}^{\mathrm{2}} \left(\mathrm{1}+{s}^{\mathrm{2}} \right)}{\mathrm{2}{u}^{\mathrm{2}} }\:\:\:\:\:\:\:\:\:….\left({i}\right) \\ $$$$−{H}={nD}−\frac{{gD}^{\mathrm{2}} \left(\mathrm{1}+{n}^{\mathrm{2}} \right)}{\mathrm{2}{v}_{\mathrm{2}} ^{\mathrm{2}} }\:\:\:\:\:\:…\left({ii}\right) \\ $$$$\:{mD}={h}+{H}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:…\left({iii}\right) \\ $$$${v}_{\mathrm{1}} ^{\mathrm{2}} ={u}^{\mathrm{2}} −\mathrm{2}{gH}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:…\left({iv}\right) \\ $$$${v}_{\mathrm{2}} \mathrm{cos}\:\phi={ev}_{\mathrm{1}} \mathrm{cos}\:\varphi \\ $$$${v}_{\mathrm{2}} \mathrm{sin}\:\phi={v}_{\mathrm{1}} \mathrm{sin}\:\varphi \\ $$$${unknowns}\:\:{u},\:{m},\:{n},\:{D},\:{H},\:\varphi,\:\phi \\ $$$${v}_{\mathrm{2}} ^{\mathrm{2}} ={v}_{\mathrm{1}} ^{\mathrm{2}} \left({e}^{\mathrm{2}} \mathrm{cos}\:^{\mathrm{2}} \varphi+\mathrm{sin}\:^{\mathrm{2}} \varphi\right) \\ $$$$\Rightarrow\:{v}_{\mathrm{2}} ^{\mathrm{2}} =\left({u}^{\mathrm{2}} −\mathrm{2}{gH}\right)\left({e}^{\mathrm{2}} \mathrm{cos}\:^{\mathrm{2}} \varphi+\mathrm{sin}\:^{\mathrm{2}} \varphi\right) \\ $$$${e}\mathrm{tan}\:\phi=\mathrm{tan}\:\varphi \\ $$$$\mathrm{tan}\:\delta={n}=\mathrm{tan}\:\left(\mathrm{90}°−\alpha−\phi\right) \\ $$$$\Rightarrow\:{n}=\frac{\mathrm{1}−\mathrm{tan}\:\phi}{{m}+\mathrm{tan}\:\phi}=\frac{{e}−\mathrm{tan}\:\varphi}{{em}+\mathrm{tan}\:\varphi} \\ $$$$\mathrm{tan}\:\left(\mathrm{90}°−\alpha+\varphi\right)=\frac{\sqrt{{u}^{\mathrm{2}} \mathrm{sin}\:^{\mathrm{2}} \theta−\mathrm{2}{gH}}}{{u}\mathrm{cos}\:\theta} \\ $$$$\Rightarrow\:\frac{\mathrm{1}+{m}\mathrm{tan}\:\varphi}{{m}−\mathrm{tan}\:\varphi}=\sqrt{{s}^{\mathrm{2}} −\frac{\mathrm{2}{gH}\left(\mathrm{1}+{s}^{\mathrm{2}} \right)}{{u}^{\mathrm{2}} }} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\:{f}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:…..\left({v}\right) \\ $$$$\:\:\mathrm{1}+{m}\mathrm{tan}\:\varphi={mf}−{f}\mathrm{tan}\:\varphi \\ $$$$\Rightarrow\:\:\mathrm{tan}\:\varphi=\frac{{mf}−\mathrm{1}}{{m}+{f}}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:….\left({vi}\right) \\ $$$$\:\:{n}=\:\frac{{e}−\frac{{mf}−\mathrm{1}}{{m}+{f}}}{{em}+\frac{{mf}−\mathrm{1}}{{m}+{f}}}=\frac{{em}+{ef}−{mf}+\mathrm{1}}{{em}^{\mathrm{2}} +{emf}+{mf}−\mathrm{1}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:….\left({vii}\right) \\ $$$${from}\:\left({ii}\right)\:\:{which}\:{is} \\ $$$$−{H}={nD}−\frac{{gD}^{\mathrm{2}} \left(\mathrm{1}+{n}^{\mathrm{2}} \right)}{\mathrm{2}{v}_{\mathrm{2}} ^{\mathrm{2}} } \\ $$$$\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_ \\ $$$$−{H}={nD}−\frac{{gD}^{\mathrm{2}} \left(\mathrm{1}+{n}^{\mathrm{2}} \right)\left\{\left({mf}−\mathrm{1}\right)^{\mathrm{2}} +\left({m}+{f}\right)^{\mathrm{2}} \right\}}{\mathrm{2}\left({u}^{\mathrm{2}} −\mathrm{2}{gH}\right)\left[{e}^{\mathrm{2}} +\left({mf}−\mathrm{1}\right)^{\mathrm{2}} \right]} \\ $$$$−{H}=\left(\frac{{em}+{ef}−{mf}+\mathrm{1}}{{em}^{\mathrm{2}} +{emf}+{mf}−\mathrm{1}}\right){D}−\frac{{gD}^{\mathrm{2}} \left\{\mathrm{1}+\left(\frac{{em}+{ef}−{mf}+\mathrm{1}}{{em}^{\mathrm{2}} +{emf}+{mf}−\mathrm{1}}\right)^{\mathrm{2}} \right\}\left\{\left({mf}−\mathrm{1}\right)^{\mathrm{2}} +\left({m}+{f}\right)^{\mathrm{2}} \right\}}{\mathrm{2}\left({u}^{\mathrm{2}} −\mathrm{2}{gH}\right)\left[{e}^{\mathrm{2}} +\left({mf}−\mathrm{1}\right)^{\mathrm{2}} \right]}\:\:\:\:\:\:\: \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:…\left(\mathrm{1}\right) \\ $$$$ \\ $$$${where}\:{f}=\sqrt{{s}^{\mathrm{2}} −\frac{\mathrm{2}{gH}\left(\mathrm{1}+{s}^{\mathrm{2}} \right)}{{u}^{\mathrm{2}} }} \\ $$$$\:{mD}={h}+{H}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\left(\mathrm{2}\right)\:…\left({iii}\right) \\ $$$${H}={sD}−\frac{{gD}^{\mathrm{2}} \left(\mathrm{1}+{s}^{\mathrm{2}} \right)}{\mathrm{2}{u}^{\mathrm{2}} }\:\:\:\:\:\:\:\:\left(\mathrm{3}\right)\:….\left({i}\right) \\ $$$${Now}\:{unknowns}\:{are}\:{u},\:{s},\:{H},\:{D} \\ $$$${From}\:\:\left(\mathrm{1}\right),\:\left(\mathrm{2}\right),\:\left(\mathrm{3}\right)\:{we}\:{obtain} \\ $$$${a}\:{correlation}\:{in}\:{u}\:{and}\:{s}=\mathrm{tan}\:\theta \\ $$$$………..\blacksquare \\ $$
Commented by mr W last updated on 04/Nov/19
$${correct}\:{way}\:{sir}! \\ $$
Answered by mr W last updated on 04/Nov/19
Commented by mr W last updated on 04/Nov/19
![AB=s v_1 sin ϕ=v_2 sin φ ev_1 cos ϕ=v_2 cos φ ((tan ϕ)/e)=tan φ ⇒sin φ=((tan ϕ)/( (√(e^2 +tan^2 ϕ))))=(1/( (√(1+(e^2 /(tan^2 ϕ)))))) ⇒φ=tan^(−1) ((tan ϕ)/e) ⇒v_2 =((sin ϕ)/(sin φ))v=(√(sin^2 ϕ+e^2 cos^2 ϕ)) v t=((s cos α)/(v cos ((π/2)−α+ϕ)))=((s cos α)/(v sin (α−ϕ))) −s sin α+h=v cos (α−ϕ) t−((gt^2 )/2) ((gs^2 cos^2 α)/(2v^2 ))[1+(1/(tan^2 (α−ϕ)))]−[tan α+(1/(tan (α−ϕ)))]s cos α+h=0 ⇒((s cos α)/(2h))=(1/(((2gh)/v^2 )[1+(1/(tan^2 (α−ϕ)))])){tan α+(1/(tan (α−ϕ)))+(√([tan α+(1/(tan (α−ϕ)))]^2 −((2gh)/v^2 )[1+(1/(tan^2 (α−ϕ)))]))} with V=((2gh)/v^2 ) ⇒((s cos α)/(2h))=(1/(V[1+(1/(tan^2 (α−ϕ)))])){tan α+(1/(tan (α−ϕ)))+(√([tan α+(1/(tan (α−ϕ)))]^2 −V[1+(1/(tan^2 (α−ϕ)))]))} t=((s cos α)/(v_2 cos ((π/2)−α−φ)))=((s cos α)/(v sin (α+φ)(√(sin^2 ϕ+e^2 cos^2 ϕ)))) −s sin α+h=v_2 cos (α+φ) t−((gt^2 )/2) −s sin α+h=((s cos α)/(tan (α+φ)))−((gs^2 cos^2 α)/(2v^2 (sin^2 ϕ+e^2 cos^2 ϕ)))[1+(1/(tan^2 (α+φ)))] ((gs^2 cos^2 α)/(2v^2 (sin^2 ϕ+e^2 cos^2 ϕ)))[1+(1/(tan^2 (α+φ)))]−[tan α+(1/(tan (α+φ)))]s cos α+h=0 ⇒((s cos α)/(2h))=(1/(((2gh)/(v^2 (sin^2 ϕ+e^2 cos^2 ϕ)))[1+(1/(tan^2 (α+tan^(−1) ((tan ϕ)/e))))])){tan α+(1/(tan (α+tan^(−1) ((tan ϕ)/e))))+(√([tan α+(1/(tan (α+tan^(−1) ((tan ϕ)/e))))]^2 −((2gh)/(v^2 (sin^2 ϕ+e^2 cos^2 ϕ)))[1+(1/(tan^2 (α+tan^(−1) ((tan ϕ)/e))))]))} ⇒((s cos α)/(2h))=(1/((V/((sin^2 ϕ+e^2 cos^2 ϕ)))[1+(1/(tan^2 (α+tan^(−1) ((tan ϕ)/e))))])){tan α+(1/(tan (α+tan^(−1) ((tan ϕ)/e))))+(√([tan α+(1/(tan (α+tan^(−1) ((tan ϕ)/e))))]^2 −(V/((sin^2 ϕ+e^2 cos^2 ϕ)))[1+(1/(tan^2 (α+tan^(−1) ((tan ϕ)/e))))]))} (1/((V/((sin^2 ϕ+e^2 cos^2 ϕ)))[1+(1/(tan^2 (α+tan^(−1) ((tan ϕ)/e))))])){tan α+(1/(tan (α+tan^(−1) ((tan ϕ)/e))))+(√([tan α+(1/(tan (α+tan^(−1) ((tan ϕ)/e))))]^2 −(V/((sin^2 ϕ+e^2 cos^2 ϕ)))[1+(1/(tan^2 (α+tan^(−1) ((tan ϕ)/e))))]))}=(1/(V[1+(1/(tan^2 (α−ϕ)))])){tan α+(1/(tan (α−ϕ)))+(√([tan α+(1/(tan (α−ϕ)))]^2 −V[1+(1/(tan^2 (α−ϕ)))]))} ⇒((tan α+(1/(tan (α−ϕ)))+(√([tan α+(1/(tan (α−ϕ)))]^2 −V[1+(1/(tan^2 (α−ϕ)))])))/(tan α+(1/(tan (α+tan^(−1) ((tan ϕ)/e))))+(√([tan α+(1/(tan (α+tan^(−1) ((tan ϕ)/e))))]^2 −(V/((sin^2 ϕ+e^2 cos^2 ϕ)))[1+(1/(tan^2 (α+tan^(−1) ((tan ϕ)/e))))]))))=(((sin^2 ϕ+e^2 cos^2 ϕ)[1+(1/(tan^2 (α−ϕ)))])/([1+(1/(tan^2 (α+tan^(−1) ((tan ϕ)/e))))])) ...(i) u^2 =v^2 +2g(s sin α−h) with U=((2gh)/u^2 ) ((1/U)+1)V=1+((2 tan α)/(1+(1/(tan^2 (α−ϕ))))){tan α+(1/(tan (α−ϕ)))+(√([tan α+(1/(tan (α−ϕ)))]^2 −V[1+(1/(tan^2 (α−ϕ)))]))}](https://www.tinkutara.com/question/Q72911.png)
$${AB}={s} \\ $$$${v}_{\mathrm{1}} \mathrm{sin}\:\varphi={v}_{\mathrm{2}} \mathrm{sin}\:\phi \\ $$$${ev}_{\mathrm{1}} \mathrm{cos}\:\varphi={v}_{\mathrm{2}} \mathrm{cos}\:\phi \\ $$$$\frac{\mathrm{tan}\:\varphi}{{e}}=\mathrm{tan}\:\phi\:\Rightarrow\mathrm{sin}\:\phi=\frac{\mathrm{tan}\:\varphi}{\:\sqrt{{e}^{\mathrm{2}} +\mathrm{tan}^{\mathrm{2}} \:\varphi}}=\frac{\mathrm{1}}{\:\sqrt{\mathrm{1}+\frac{{e}^{\mathrm{2}} }{\mathrm{tan}^{\mathrm{2}} \:\varphi}}} \\ $$$$\Rightarrow\phi=\mathrm{tan}^{−\mathrm{1}} \frac{\mathrm{tan}\:\varphi}{{e}} \\ $$$$\Rightarrow{v}_{\mathrm{2}} =\frac{\mathrm{sin}\:\varphi}{\mathrm{sin}\:\phi}{v}=\sqrt{\mathrm{sin}^{\mathrm{2}} \:\varphi+{e}^{\mathrm{2}} \mathrm{cos}^{\mathrm{2}} \:\varphi}\:{v} \\ $$$${t}=\frac{{s}\:\mathrm{cos}\:\alpha}{{v}\:\mathrm{cos}\:\left(\frac{\pi}{\mathrm{2}}−\alpha+\varphi\right)}=\frac{{s}\:\mathrm{cos}\:\alpha}{{v}\:\mathrm{sin}\:\left(\alpha−\varphi\right)} \\ $$$$−{s}\:\mathrm{sin}\:\alpha+{h}={v}\:\mathrm{cos}\:\left(\alpha−\varphi\right)\:{t}−\frac{{gt}^{\mathrm{2}} }{\mathrm{2}} \\ $$$$\frac{{gs}^{\mathrm{2}} \mathrm{cos}^{\mathrm{2}} \:\alpha}{\mathrm{2}{v}^{\mathrm{2}} }\left[\mathrm{1}+\frac{\mathrm{1}}{\mathrm{tan}^{\mathrm{2}} \:\left(\alpha−\varphi\right)}\right]−\left[\mathrm{tan}\:\alpha+\frac{\mathrm{1}}{\mathrm{tan}\:\left(\alpha−\varphi\right)}\right]{s}\:\mathrm{cos}\:\alpha+{h}=\mathrm{0} \\ $$$$\Rightarrow\frac{{s}\:\mathrm{cos}\:\alpha}{\mathrm{2}{h}}=\frac{\mathrm{1}}{\frac{\mathrm{2}{gh}}{{v}^{\mathrm{2}} }\left[\mathrm{1}+\frac{\mathrm{1}}{\mathrm{tan}^{\mathrm{2}} \:\left(\alpha−\varphi\right)}\right]}\left\{\mathrm{tan}\:\alpha+\frac{\mathrm{1}}{\mathrm{tan}\:\left(\alpha−\varphi\right)}+\sqrt{\left[\mathrm{tan}\:\alpha+\frac{\mathrm{1}}{\mathrm{tan}\:\left(\alpha−\varphi\right)}\right]^{\mathrm{2}} −\frac{\mathrm{2}{gh}}{{v}^{\mathrm{2}} }\left[\mathrm{1}+\frac{\mathrm{1}}{\mathrm{tan}^{\mathrm{2}} \:\left(\alpha−\varphi\right)}\right]}\right\} \\ $$$${with}\:{V}=\frac{\mathrm{2}{gh}}{{v}^{\mathrm{2}} } \\ $$$$\Rightarrow\frac{{s}\:\mathrm{cos}\:\alpha}{\mathrm{2}{h}}=\frac{\mathrm{1}}{{V}\left[\mathrm{1}+\frac{\mathrm{1}}{\mathrm{tan}^{\mathrm{2}} \:\left(\alpha−\varphi\right)}\right]}\left\{\mathrm{tan}\:\alpha+\frac{\mathrm{1}}{\mathrm{tan}\:\left(\alpha−\varphi\right)}+\sqrt{\left[\mathrm{tan}\:\alpha+\frac{\mathrm{1}}{\mathrm{tan}\:\left(\alpha−\varphi\right)}\right]^{\mathrm{2}} −{V}\left[\mathrm{1}+\frac{\mathrm{1}}{\mathrm{tan}^{\mathrm{2}} \:\left(\alpha−\varphi\right)}\right]}\right\} \\ $$$${t}=\frac{{s}\:\mathrm{cos}\:\alpha}{{v}_{\mathrm{2}} \:\mathrm{cos}\:\left(\frac{\pi}{\mathrm{2}}−\alpha−\phi\right)}=\frac{{s}\:\mathrm{cos}\:\alpha}{{v}\:\mathrm{sin}\:\left(\alpha+\phi\right)\sqrt{\mathrm{sin}^{\mathrm{2}} \:\varphi+{e}^{\mathrm{2}} \mathrm{cos}^{\mathrm{2}} \:\varphi}} \\ $$$$−{s}\:\mathrm{sin}\:\alpha+{h}={v}_{\mathrm{2}} \:\mathrm{cos}\:\left(\alpha+\phi\right)\:{t}−\frac{{gt}^{\mathrm{2}} }{\mathrm{2}} \\ $$$$−{s}\:\mathrm{sin}\:\alpha+{h}=\frac{{s}\:\mathrm{cos}\:\alpha}{\mathrm{tan}\:\left(\alpha+\phi\right)}−\frac{{gs}^{\mathrm{2}} \:\mathrm{cos}^{\mathrm{2}} \:\alpha}{\mathrm{2}{v}^{\mathrm{2}} \left(\mathrm{sin}^{\mathrm{2}} \:\varphi+{e}^{\mathrm{2}} \mathrm{cos}^{\mathrm{2}} \:\varphi\right)}\left[\mathrm{1}+\frac{\mathrm{1}}{\mathrm{tan}^{\mathrm{2}} \:\left(\alpha+\phi\right)}\right] \\ $$$$\frac{{gs}^{\mathrm{2}} \:\mathrm{cos}^{\mathrm{2}} \:\alpha}{\mathrm{2}{v}^{\mathrm{2}} \left(\mathrm{sin}^{\mathrm{2}} \:\varphi+{e}^{\mathrm{2}} \mathrm{cos}^{\mathrm{2}} \:\varphi\right)}\left[\mathrm{1}+\frac{\mathrm{1}}{\mathrm{tan}^{\mathrm{2}} \:\left(\alpha+\phi\right)}\right]−\left[\mathrm{tan}\:\alpha+\frac{\mathrm{1}}{\mathrm{tan}\:\left(\alpha+\phi\right)}\right]{s}\:\mathrm{cos}\:\alpha+{h}=\mathrm{0} \\ $$$$\Rightarrow\frac{{s}\:\mathrm{cos}\:\alpha}{\mathrm{2}{h}}=\frac{\mathrm{1}}{\frac{\mathrm{2}{gh}}{{v}^{\mathrm{2}} \left(\mathrm{sin}^{\mathrm{2}} \:\varphi+{e}^{\mathrm{2}} \mathrm{cos}^{\mathrm{2}} \:\varphi\right)}\left[\mathrm{1}+\frac{\mathrm{1}}{\mathrm{tan}^{\mathrm{2}} \:\left(\alpha+\mathrm{tan}^{−\mathrm{1}} \frac{\mathrm{tan}\:\varphi}{{e}}\right)}\right]}\left\{\mathrm{tan}\:\alpha+\frac{\mathrm{1}}{\mathrm{tan}\:\left(\alpha+\mathrm{tan}^{−\mathrm{1}} \frac{\mathrm{tan}\:\varphi}{{e}}\right)}+\sqrt{\left[\mathrm{tan}\:\alpha+\frac{\mathrm{1}}{\mathrm{tan}\:\left(\alpha+\mathrm{tan}^{−\mathrm{1}} \frac{\mathrm{tan}\:\varphi}{{e}}\right)}\right]^{\mathrm{2}} −\frac{\mathrm{2}{gh}}{{v}^{\mathrm{2}} \left(\mathrm{sin}^{\mathrm{2}} \:\varphi+{e}^{\mathrm{2}} \mathrm{cos}^{\mathrm{2}} \:\varphi\right)}\left[\mathrm{1}+\frac{\mathrm{1}}{\mathrm{tan}^{\mathrm{2}} \:\left(\alpha+\mathrm{tan}^{−\mathrm{1}} \frac{\mathrm{tan}\:\varphi}{{e}}\right)}\right]}\right\} \\ $$$$\Rightarrow\frac{{s}\:\mathrm{cos}\:\alpha}{\mathrm{2}{h}}=\frac{\mathrm{1}}{\frac{{V}}{\left(\mathrm{sin}^{\mathrm{2}} \:\varphi+{e}^{\mathrm{2}} \mathrm{cos}^{\mathrm{2}} \:\varphi\right)}\left[\mathrm{1}+\frac{\mathrm{1}}{\mathrm{tan}^{\mathrm{2}} \:\left(\alpha+\mathrm{tan}^{−\mathrm{1}} \frac{\mathrm{tan}\:\varphi}{{e}}\right)}\right]}\left\{\mathrm{tan}\:\alpha+\frac{\mathrm{1}}{\mathrm{tan}\:\left(\alpha+\mathrm{tan}^{−\mathrm{1}} \frac{\mathrm{tan}\:\varphi}{{e}}\right)}+\sqrt{\left[\mathrm{tan}\:\alpha+\frac{\mathrm{1}}{\mathrm{tan}\:\left(\alpha+\mathrm{tan}^{−\mathrm{1}} \frac{\mathrm{tan}\:\varphi}{{e}}\right)}\right]^{\mathrm{2}} −\frac{{V}}{\left(\mathrm{sin}^{\mathrm{2}} \:\varphi+{e}^{\mathrm{2}} \mathrm{cos}^{\mathrm{2}} \:\varphi\right)}\left[\mathrm{1}+\frac{\mathrm{1}}{\mathrm{tan}^{\mathrm{2}} \:\left(\alpha+\mathrm{tan}^{−\mathrm{1}} \frac{\mathrm{tan}\:\varphi}{{e}}\right)}\right]}\right\} \\ $$$$\frac{\mathrm{1}}{\frac{{V}}{\left(\mathrm{sin}^{\mathrm{2}} \:\varphi+{e}^{\mathrm{2}} \mathrm{cos}^{\mathrm{2}} \:\varphi\right)}\left[\mathrm{1}+\frac{\mathrm{1}}{\mathrm{tan}^{\mathrm{2}} \:\left(\alpha+\mathrm{tan}^{−\mathrm{1}} \frac{\mathrm{tan}\:\varphi}{{e}}\right)}\right]}\left\{\mathrm{tan}\:\alpha+\frac{\mathrm{1}}{\mathrm{tan}\:\left(\alpha+\mathrm{tan}^{−\mathrm{1}} \frac{\mathrm{tan}\:\varphi}{{e}}\right)}+\sqrt{\left[\mathrm{tan}\:\alpha+\frac{\mathrm{1}}{\mathrm{tan}\:\left(\alpha+\mathrm{tan}^{−\mathrm{1}} \frac{\mathrm{tan}\:\varphi}{{e}}\right)}\right]^{\mathrm{2}} −\frac{{V}}{\left(\mathrm{sin}^{\mathrm{2}} \:\varphi+{e}^{\mathrm{2}} \mathrm{cos}^{\mathrm{2}} \:\varphi\right)}\left[\mathrm{1}+\frac{\mathrm{1}}{\mathrm{tan}^{\mathrm{2}} \:\left(\alpha+\mathrm{tan}^{−\mathrm{1}} \frac{\mathrm{tan}\:\varphi}{{e}}\right)}\right]}\right\}=\frac{\mathrm{1}}{{V}\left[\mathrm{1}+\frac{\mathrm{1}}{\mathrm{tan}^{\mathrm{2}} \:\left(\alpha−\varphi\right)}\right]}\left\{\mathrm{tan}\:\alpha+\frac{\mathrm{1}}{\mathrm{tan}\:\left(\alpha−\varphi\right)}+\sqrt{\left[\mathrm{tan}\:\alpha+\frac{\mathrm{1}}{\mathrm{tan}\:\left(\alpha−\varphi\right)}\right]^{\mathrm{2}} −{V}\left[\mathrm{1}+\frac{\mathrm{1}}{\mathrm{tan}^{\mathrm{2}} \:\left(\alpha−\varphi\right)}\right]}\right\} \\ $$$$\Rightarrow\frac{\mathrm{tan}\:\alpha+\frac{\mathrm{1}}{\mathrm{tan}\:\left(\alpha−\varphi\right)}+\sqrt{\left[\mathrm{tan}\:\alpha+\frac{\mathrm{1}}{\mathrm{tan}\:\left(\alpha−\varphi\right)}\right]^{\mathrm{2}} −{V}\left[\mathrm{1}+\frac{\mathrm{1}}{\mathrm{tan}^{\mathrm{2}} \:\left(\alpha−\varphi\right)}\right]}}{\mathrm{tan}\:\alpha+\frac{\mathrm{1}}{\mathrm{tan}\:\left(\alpha+\mathrm{tan}^{−\mathrm{1}} \frac{\mathrm{tan}\:\varphi}{{e}}\right)}+\sqrt{\left[\mathrm{tan}\:\alpha+\frac{\mathrm{1}}{\mathrm{tan}\:\left(\alpha+\mathrm{tan}^{−\mathrm{1}} \frac{\mathrm{tan}\:\varphi}{{e}}\right)}\right]^{\mathrm{2}} −\frac{{V}}{\left(\mathrm{sin}^{\mathrm{2}} \:\varphi+{e}^{\mathrm{2}} \mathrm{cos}^{\mathrm{2}} \:\varphi\right)}\left[\mathrm{1}+\frac{\mathrm{1}}{\mathrm{tan}^{\mathrm{2}} \:\left(\alpha+\mathrm{tan}^{−\mathrm{1}} \frac{\mathrm{tan}\:\varphi}{{e}}\right)}\right]}}=\frac{\left(\mathrm{sin}^{\mathrm{2}} \:\varphi+{e}^{\mathrm{2}} \mathrm{cos}^{\mathrm{2}} \:\varphi\right)\left[\mathrm{1}+\frac{\mathrm{1}}{\mathrm{tan}^{\mathrm{2}} \:\left(\alpha−\varphi\right)}\right]}{\left[\mathrm{1}+\frac{\mathrm{1}}{\mathrm{tan}^{\mathrm{2}} \:\left(\alpha+\mathrm{tan}^{−\mathrm{1}} \frac{\mathrm{tan}\:\varphi}{{e}}\right)}\right]}\:\:\:\:…\left({i}\right) \\ $$$${u}^{\mathrm{2}} ={v}^{\mathrm{2}} +\mathrm{2}{g}\left({s}\:\mathrm{sin}\:\alpha−{h}\right) \\ $$$${with}\:{U}=\frac{\mathrm{2}{gh}}{{u}^{\mathrm{2}} } \\ $$$$\left(\frac{\mathrm{1}}{{U}}+\mathrm{1}\right){V}=\mathrm{1}+\frac{\mathrm{2}\:\mathrm{tan}\:\alpha}{\mathrm{1}+\frac{\mathrm{1}}{\mathrm{tan}^{\mathrm{2}} \:\left(\alpha−\varphi\right)}}\left\{\mathrm{tan}\:\alpha+\frac{\mathrm{1}}{\mathrm{tan}\:\left(\alpha−\varphi\right)}+\sqrt{\left[\mathrm{tan}\:\alpha+\frac{\mathrm{1}}{\mathrm{tan}\:\left(\alpha−\varphi\right)}\right]^{\mathrm{2}} −{V}\left[\mathrm{1}+\frac{\mathrm{1}}{\mathrm{tan}^{\mathrm{2}} \:\left(\alpha−\varphi\right)}\right]}\right\} \\ $$
Commented by mr W last updated on 04/Nov/19
$${by}\:{solving}\:{this}\:{quadratic}\:{eqn}.\:{for}\:{V} \\ $$$${we}\:{get}\:{V}={f}\left({U}\right)\:{and}\:{put}\:{it}\:{into}\:\left({i}\right)\:{to} \\ $$$${have}\:{an}\:{eqn}.\:{with}\:{U}\:{and}\:\varphi. \\ $$