Question Number 225758 by mr W last updated on 09/Nov/25
Commented by mr W last updated on 09/Nov/25
Answered by ajfour last updated on 09/Nov/25
$${v}={ds}/{dt}={u}\mathrm{cos}\:\theta_{\mathrm{0}} \mathrm{sec}\:\phi \\ $$$${dv}={u}\mathrm{cos}\:\theta_{\mathrm{0}} \mathrm{sec}\:\phi\mathrm{tan}\:\phi{d}\phi \\ $$$${g}\mathrm{sin}\:\phi{ds}=−{vdv} \\ $$$$\int{ds}=−\int\frac{{vdv}}{{g}\mathrm{sin}\:\phi}=−\frac{{u}^{\mathrm{2}} \mathrm{cos}\:^{\mathrm{2}} \theta_{\mathrm{0}} }{{g}}\int_{\:\theta_{\mathrm{0}} } ^{\:−\theta_{\mathrm{0}} } \mathrm{sec}\:^{\mathrm{3}} \phi{d}\phi \\ $$$$\:{s}\left(\theta_{\mathrm{0}} \right)=\frac{\mathrm{2}{u}^{\mathrm{2}} \mathrm{cos}\:^{\mathrm{2}} \theta_{\mathrm{0}} }{{g}}\left\{\frac{\mathrm{sin}\:\theta_{\mathrm{0}} }{\mathrm{cos}\:^{\mathrm{2}} \theta_{\mathrm{0}} }+\frac{\mathrm{1}}{\mathrm{2}}\mathrm{ln}\:\mid\frac{\mathrm{1}+\mathrm{sin}\:\theta_{\mathrm{0}} }{\mathrm{1}−\mathrm{sin}\:\theta_{\mathrm{0}} }\mid\right\} \\ $$$$ \\ $$
Commented by mr W last updated on 09/Nov/25
$${thanks}\:{sir}! \\ $$$${do}\:{you}\:{also}\:{get}\:\theta_{\mathrm{0}} =\mathrm{56}.\mathrm{46}°? \\ $$
Answered by taha3738 last updated on 09/Nov/25
![L(θ)=∫_0 ^( T) (√((v_x (t))^2 +(v_y (t))^2 ))dt =∫_0 ^( T) (√((v_0 cos θ)^2 +(v_0 sin θ−gt)^2 ))dt =∫_0 ^( T) (√(g^2 t^2 −2gv sin θt + v_0 ^2 )) dt =∫_0 ^( T) (√(g^2 (t^2 −((2v_0 sin θ)/g))+v_0 ^2 )) dt =∫_0 ^T (√(g^2 [(t−((2v_0 sin θ)/g))^2 −(((v_0 sin θ)/g))^2 ]+v_0 ^2 )) dt =∫_0 ^( T) (√(g^2 (t−((2v_0 sin θ)/g))^2 +v_0 ^2 cos^2 θ)) dt u=t−((2v_0 sin θ)/g) , a=((v_0 sin θ)/g) , du=dt = ∫_(−a) ^( a) (√(g^2 u^2 +v_0 ^2 cos^2 θ)) du There′s a formula for ∫ (√(p^2 u^2 +q^2 )) du](https://www.tinkutara.com/question/Q225762.png)
$${L}\left(\theta\right)=\int_{\mathrm{0}} ^{\:{T}} \:\sqrt{\left({v}_{{x}} \left({t}\right)\right)^{\mathrm{2}} +\left({v}_{{y}} \left({t}\right)\right)^{\mathrm{2}} }{dt} \\ $$$$=\int_{\mathrm{0}} ^{\:{T}} \sqrt{\left({v}_{\mathrm{0}} \mathrm{cos}\:\theta\right)^{\mathrm{2}} +\left({v}_{\mathrm{0}} \mathrm{sin}\:\theta−{gt}\right)^{\mathrm{2}} }{dt} \\ $$$$=\int_{\mathrm{0}} ^{\:{T}} \sqrt{{g}^{\mathrm{2}} {t}^{\mathrm{2}} −\mathrm{2}{gv}\:\mathrm{sin}\:\theta{t}\:+\:{v}_{\mathrm{0}} ^{\mathrm{2}} }\:{dt} \\ $$$$=\int_{\mathrm{0}} ^{\:{T}} \:\sqrt{{g}^{\mathrm{2}} \left({t}^{\mathrm{2}} −\frac{\mathrm{2}{v}_{\mathrm{0}} \mathrm{sin}\:\theta}{{g}}\right)+{v}_{\mathrm{0}} ^{\mathrm{2}} }\:{dt} \\ $$$$=\int_{\mathrm{0}} ^{{T}} \:\sqrt{{g}^{\mathrm{2}} \left[\left({t}−\frac{\mathrm{2}{v}_{\mathrm{0}} \mathrm{sin}\:\theta}{{g}}\right)^{\mathrm{2}} −\left(\frac{{v}_{\mathrm{0}} \mathrm{sin}\:\theta}{{g}}\right)^{\mathrm{2}} \right]+{v}_{\mathrm{0}} ^{\mathrm{2}} }\:{dt} \\ $$$$=\int_{\mathrm{0}} ^{\:{T}} \:\sqrt{{g}^{\mathrm{2}} \left({t}−\frac{\mathrm{2}{v}_{\mathrm{0}} \mathrm{sin}\:\theta}{{g}}\right)^{\mathrm{2}} +{v}_{\mathrm{0}} ^{\mathrm{2}} \mathrm{cos}^{\mathrm{2}} \theta}\:{dt} \\ $$$${u}={t}−\frac{\mathrm{2}{v}_{\mathrm{0}} \mathrm{sin}\:\theta}{{g}}\:,\:{a}=\frac{{v}_{\mathrm{0}} \mathrm{sin}\:\theta}{{g}}\:,\:{du}={dt} \\ $$$$=\:\int_{−{a}} ^{\:{a}} \:\sqrt{{g}^{\mathrm{2}} {u}^{\mathrm{2}} +{v}_{\mathrm{0}} ^{\mathrm{2}} \mathrm{cos}^{\mathrm{2}} \:\theta}\:{du} \\ $$$${There}'{s}\:{a}\:{formula}\:{for}\:\int\:\sqrt{{p}^{\mathrm{2}} {u}^{\mathrm{2}} +{q}^{\mathrm{2}} }\:{du} \\ $$
Answered by taha3738 last updated on 09/Nov/25
![∫ (√(p^2 u^2 +q^2 )) du = (u/2)(√(p^2 u^2 +q^2 )) + (q^2 /(2p)) ln ∣pu+(√(p^2 u^2 +q^2 ))∣ + C So ∫_(−a) ^( a) (√(g^2 u^2 +v_0 ^2 cos^2 θ)) du =2[(u/2)(√(g^2 u^2 +v_0 ^2 cos^2 θ)) + ((v_0 ^2 cos^2 θ)/(2g)) ln(gu+(√(g^2 u^2 +v_0 ^2 cos^2 θ)))]_0 ^a The algebra gets really messy but it gives : L(θ)=(v_0 ^2 /g)(sin θ + cos^2 θ ln(((1+sin θ)/(cos θ)))) (dL/dθ) =(v_0 ^2 /g)[cos θ −2ln(((1+sin θ)/(cos θ)))sin θcosθ+ ((1+sin θ)/(cos^2 θ))×((cos θ)/(1+sin θ))×cos^2 θ] (dL/dθ) = (v_0 ^2 /g)[2cos θ − 2 cos θ sin θ ln (((1+sin θ)/(cos θ)))] Set (dL/dθ) = 0 to get the condition : sin θ ln (((1+sin θ)/(cos θ))) = 1 numerically : θ≈56.47°](https://www.tinkutara.com/question/Q225763.png)
$$\int\:\sqrt{{p}^{\mathrm{2}} {u}^{\mathrm{2}} +{q}^{\mathrm{2}} }\:{du}\:=\:\frac{{u}}{\mathrm{2}}\sqrt{{p}^{\mathrm{2}} {u}^{\mathrm{2}} +{q}^{\mathrm{2}} }\:+\:\frac{{q}^{\mathrm{2}} }{\mathrm{2}{p}}\:\mathrm{ln}\:\mid{pu}+\sqrt{{p}^{\mathrm{2}} {u}^{\mathrm{2}} +{q}^{\mathrm{2}} }\mid\:+\:{C} \\ $$$${So}\: \\ $$$$\int_{−{a}} ^{\:{a}} \:\sqrt{{g}^{\mathrm{2}} {u}^{\mathrm{2}} +{v}_{\mathrm{0}} ^{\mathrm{2}} \mathrm{cos}^{\mathrm{2}} \:\theta}\:{du}\: \\ $$$$=\mathrm{2}\left[\frac{{u}}{\mathrm{2}}\sqrt{{g}^{\mathrm{2}} {u}^{\mathrm{2}} +{v}_{\mathrm{0}} ^{\mathrm{2}} \mathrm{cos}^{\mathrm{2}} \theta}\:+\:\frac{{v}_{\mathrm{0}} ^{\mathrm{2}} \mathrm{cos}^{\mathrm{2}} \theta}{\mathrm{2}{g}}\:\mathrm{ln}\left({gu}+\sqrt{{g}^{\mathrm{2}} {u}^{\mathrm{2}} +{v}_{\mathrm{0}} ^{\mathrm{2}} \mathrm{cos}^{\mathrm{2}} \theta}\right)\right]_{\mathrm{0}} ^{{a}} \\ $$$${The}\:{algebra}\:{gets}\:{really}\:{messy}\:{but}\:{it}\:{gives}\:: \\ $$$${L}\left(\theta\right)=\frac{{v}_{\mathrm{0}} ^{\mathrm{2}} }{{g}}\left(\mathrm{sin}\:\theta\:+\:\mathrm{cos}^{\mathrm{2}} \theta\:\mathrm{ln}\left(\frac{\mathrm{1}+\mathrm{sin}\:\theta}{\mathrm{cos}\:\theta}\right)\right) \\ $$$$\frac{{dL}}{{d}\theta}\:=\frac{{v}_{\mathrm{0}} ^{\mathrm{2}} }{{g}}\left[\mathrm{cos}\:\theta\:−\mathrm{2ln}\left(\frac{\mathrm{1}+\mathrm{sin}\:\theta}{\mathrm{cos}\:\theta}\right)\mathrm{sin}\:\theta\mathrm{cos}\theta+\:\frac{\mathrm{1}+\mathrm{sin}\:\theta}{\mathrm{cos}^{\mathrm{2}} \theta}×\frac{\mathrm{cos}\:\theta}{\mathrm{1}+\mathrm{sin}\:\theta}×\mathrm{cos}^{\mathrm{2}} \theta\right] \\ $$$$\frac{{dL}}{{d}\theta}\:=\:\frac{{v}_{\mathrm{0}} ^{\mathrm{2}} }{{g}}\left[\mathrm{2cos}\:\theta\:−\:\mathrm{2}\:\mathrm{cos}\:\theta\:\mathrm{sin}\:\theta\:\mathrm{ln}\:\left(\frac{\mathrm{1}+\mathrm{sin}\:\theta}{\mathrm{cos}\:\theta}\right)\right] \\ $$$${Set}\:\frac{{dL}}{{d}\theta}\:=\:\mathrm{0}\:{to}\:{get}\:{the}\:{condition}\:: \\ $$$$\mathrm{sin}\:\theta\:\mathrm{ln}\:\left(\frac{\mathrm{1}+\mathrm{sin}\:\theta}{\mathrm{cos}\:\theta}\right)\:=\:\mathrm{1}\: \\ $$$${numerically}\::\:\theta\approx\mathrm{56}.\mathrm{47}° \\ $$
Commented by mr W last updated on 09/Nov/25
$${great}!\:{thanks}. \\ $$$${i}\:{got}\:{the}\:{same},\:{but}\:{without}\:{to}\:{find} \\ $$$${out}\:{the}\:{formula}\:{for}\:{L}. \\ $$
Answered by mr W last updated on 11/Nov/25
![x=u cos θ t ⇒dx=u cos θ dt y=u sin θ t−((gt^2 )/2) ⇒dy=(u sin θ−gt)dt 0≤t≤((2u sin θ)/g) dL=(√((dx)^2 +(dy)^2 ))=(√((u cos θ)^2 +(u sin θ−gt)^2 )) dt L=2∫_0 ^((u sin θ)/g) (√((u cos θ)^2 +(u sin θ−gt)^2 )) dt L=−((2u^2 )/g)∫_0 ^((u sin θ)/g) (√((cos θ)^2 +(sin θ−((gt)/u))^2 )) d(sin θ−((gt)/u)) let ξ=sin θ−((gt)/u) ((gL)/(2u^2 ))= =∫_0 ^(sin θ) (√(cos^2 θ+ξ^2 )) dξ L_(max) ⇒ _(max) ⇒(d /dθ)=0 (d /dθ)=cos θ−∫_0 ^(sin θ) ((cos θ sin θ dξ)/( (√(cos^2 θ+ξ^2 ))))=0 1−sin θ ∫_0 ^(sin θ) (dξ/( (√(cos^2 θ+ξ^2 ))))=0 1−sin θ [sinh^(−1) (ξ/(cos θ))]_0 ^(sin θ) =0 1−sin θ sinh^(−1) ((sin θ)/(cos θ)) =0 ⇒sin θ sinh^(−1) (tan θ)=1 ⇒θ_m ≈56.465835° ((gL)/(2u^2 ))=∫_0 ^(sin θ) (√(cos^2 θ+ξ^2 )) dξ=((cos^2 θ)/2)[sinh^(−1) (ξ/(cos θ))+(ξ/(cos θ))(√(1+((ξ/(cos θ)))^2 ))]_0 ^(sin θ) ((gL)/(2u^2 ))=((cos^2 θ)/2)[sinh^(−1) tan θ+tan θ (√(1+tan^2 θ))] ((gL)/u^2 )=cos^2 θ sinh^(−1) tan θ+sin θ ⇒L_(max) =(u^2 /(g sin θ_m ))≈1.199679(u^2 /g)](https://www.tinkutara.com/question/Q225765.png)
$${x}={u}\:\mathrm{cos}\:\theta\:{t}\:\Rightarrow{dx}={u}\:\mathrm{cos}\:\theta\:{dt} \\ $$$${y}={u}\:\mathrm{sin}\:\theta\:{t}−\frac{{gt}^{\mathrm{2}} }{\mathrm{2}}\:\Rightarrow{dy}=\left({u}\:\mathrm{sin}\:\theta−{gt}\right){dt} \\ $$$$\mathrm{0}\leqslant{t}\leqslant\frac{\mathrm{2}{u}\:\mathrm{sin}\:\theta}{{g}} \\ $$$${dL}=\sqrt{\left({dx}\right)^{\mathrm{2}} +\left({dy}\right)^{\mathrm{2}} }=\sqrt{\left({u}\:\mathrm{cos}\:\theta\right)^{\mathrm{2}} +\left({u}\:\mathrm{sin}\:\theta−{gt}\right)^{\mathrm{2}} }\:{dt} \\ $$$${L}=\mathrm{2}\int_{\mathrm{0}} ^{\frac{{u}\:\mathrm{sin}\:\theta}{{g}}} \sqrt{\left({u}\:\mathrm{cos}\:\theta\right)^{\mathrm{2}} +\left({u}\:\mathrm{sin}\:\theta−{gt}\right)^{\mathrm{2}} }\:{dt} \\ $$$${L}=−\frac{\mathrm{2}{u}^{\mathrm{2}} }{{g}}\int_{\mathrm{0}} ^{\frac{{u}\:\mathrm{sin}\:\theta}{{g}}} \sqrt{\left(\mathrm{cos}\:\theta\right)^{\mathrm{2}} +\left(\mathrm{sin}\:\theta−\frac{{gt}}{{u}}\right)^{\mathrm{2}} }\:{d}\left(\mathrm{sin}\:\theta−\frac{{gt}}{{u}}\right) \\ $$$${let}\:\xi=\mathrm{sin}\:\theta−\frac{{gt}}{{u}} \\ $$$$\frac{{gL}}{\mathrm{2}{u}^{\mathrm{2}} }=\:=\int_{\mathrm{0}} ^{\mathrm{sin}\:\theta} \sqrt{\mathrm{cos}^{\mathrm{2}} \:\theta+\xi^{\mathrm{2}} }\:{d}\xi \\ $$$${L}_{{max}} \:\Rightarrow\:\:_{{max}} \Rightarrow\frac{{d}\:}{{d}\theta}=\mathrm{0} \\ $$$$\frac{{d}\:}{{d}\theta}=\mathrm{cos}\:\theta−\int_{\mathrm{0}} ^{\mathrm{sin}\:\theta} \frac{\mathrm{cos}\:\theta\:\mathrm{sin}\:\theta\:{d}\xi}{\:\sqrt{\mathrm{cos}^{\mathrm{2}} \:\theta+\xi^{\mathrm{2}} }}=\mathrm{0} \\ $$$$\mathrm{1}−\mathrm{sin}\:\theta\:\int_{\mathrm{0}} ^{\mathrm{sin}\:\theta} \frac{{d}\xi}{\:\sqrt{\mathrm{cos}^{\mathrm{2}} \:\theta+\xi^{\mathrm{2}} }}=\mathrm{0} \\ $$$$\mathrm{1}−\mathrm{sin}\:\theta\:\left[\mathrm{sinh}^{−\mathrm{1}} \:\frac{\xi}{\mathrm{cos}\:\theta}\right]_{\mathrm{0}} ^{\mathrm{sin}\:\theta} \:=\mathrm{0} \\ $$$$\mathrm{1}−\mathrm{sin}\:\theta\:\mathrm{sinh}^{−\mathrm{1}} \:\frac{\mathrm{sin}\:\theta}{\mathrm{cos}\:\theta}\:=\mathrm{0} \\ $$$$\Rightarrow\mathrm{sin}\:\theta\:\mathrm{sinh}^{−\mathrm{1}} \:\left(\mathrm{tan}\:\theta\right)=\mathrm{1} \\ $$$$\Rightarrow\theta_{{m}} \approx\mathrm{56}.\mathrm{465835}° \\ $$$$ \\ $$$$\frac{{gL}}{\mathrm{2}{u}^{\mathrm{2}} }=\int_{\mathrm{0}} ^{\mathrm{sin}\:\theta} \sqrt{\mathrm{cos}^{\mathrm{2}} \:\theta+\xi^{\mathrm{2}} }\:{d}\xi=\frac{\mathrm{cos}^{\mathrm{2}} \:\theta}{\mathrm{2}}\left[\mathrm{sinh}^{−\mathrm{1}} \:\frac{\xi}{\mathrm{cos}\:\theta}+\frac{\xi}{\mathrm{cos}\:\theta}\sqrt{\mathrm{1}+\left(\frac{\xi}{\mathrm{cos}\:\theta}\right)^{\mathrm{2}} }\right]_{\mathrm{0}} ^{\mathrm{sin}\:\theta} \\ $$$$\frac{{gL}}{\mathrm{2}{u}^{\mathrm{2}} }=\frac{\mathrm{cos}^{\mathrm{2}} \:\theta}{\mathrm{2}}\left[\mathrm{sinh}^{−\mathrm{1}} \:\mathrm{tan}\:\theta+\mathrm{tan}\:\theta\:\sqrt{\mathrm{1}+\mathrm{tan}^{\mathrm{2}} \:\theta}\right] \\ $$$$\frac{{gL}}{{u}^{\mathrm{2}} }=\mathrm{cos}^{\mathrm{2}} \:\theta\:\mathrm{sinh}^{−\mathrm{1}} \:\mathrm{tan}\:\theta+\mathrm{sin}\:\theta \\ $$$$\Rightarrow{L}_{{max}} =\frac{{u}^{\mathrm{2}} }{{g}\:\mathrm{sin}\:\theta_{{m}} }\approx\mathrm{1}.\mathrm{199679}\frac{{u}^{\mathrm{2}} }{{g}} \\ $$
Commented by ajfour last updated on 11/Nov/25
$${Great}\:{Beauty} \\ $$
Commented by mr W last updated on 12/Nov/25
