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Question Number 165457 by mr W last updated on 02/Feb/22
Commented by MJS_new last updated on 02/Feb/22
I would like you to check this: from Sir Ajfour′s scetch I assumed that the “hill” is y=−x^2 +a^2 and the intersection is at x=(1/2) because of the “e=1”−mark; I understood that the reflecting tangent has slope −1 from there I found this: for g=−10 and a=2 I get these formulas: “hill”: y=−x^2 +4 par_1 : y=−13x^2 +6x+4 par_2 : y=−((13)/(49))x^2 +(6/(49))x+((184)/(49)) is this solution ok, is it a special solution in any way? if so, I can calculate it for any given a and other values of g (i.e. −9.81) thank you!
$$\mathrm{I}\:\mathrm{would}\:\mathrm{like}\:\mathrm{you}\:\mathrm{to}\:\mathrm{check}\:\mathrm{this}: \\ $$$$\mathrm{from}\:\mathrm{Sir}\:\mathrm{Ajfour}'\mathrm{s}\:\mathrm{scetch}\:\mathrm{I}\:\mathrm{assumed}\:\mathrm{that}\:\mathrm{the} \\ $$$$“\mathrm{hill}''\:\mathrm{is}\:{y}=−{x}^{\mathrm{2}} +{a}^{\mathrm{2}} \:\mathrm{and}\:\mathrm{the}\:\mathrm{intersection}\:\mathrm{is} \\ $$$$\mathrm{at}\:{x}=\frac{\mathrm{1}}{\mathrm{2}}\:\mathrm{because}\:\mathrm{of}\:\mathrm{the}\:“{e}=\mathrm{1}''−\mathrm{mark};\:\mathrm{I} \\ $$$$\mathrm{understood}\:\mathrm{that}\:\mathrm{the}\:\mathrm{reflecting}\:\mathrm{tangent}\:\mathrm{has} \\ $$$$\mathrm{slope}\:−\mathrm{1} \\ $$$$\mathrm{from}\:\mathrm{there}\:\mathrm{I}\:\mathrm{found}\:\mathrm{this}: \\ $$$$\mathrm{for}\:{g}=−\mathrm{10}\:\mathrm{and}\:{a}=\mathrm{2}\:\mathrm{I}\:\mathrm{get}\:\mathrm{these}\:\mathrm{formulas}: \\ $$$$“\mathrm{hill}'':\:{y}=−{x}^{\mathrm{2}} +\mathrm{4} \\ $$$$\mathrm{par}_{\mathrm{1}} :\:{y}=−\mathrm{13}{x}^{\mathrm{2}} +\mathrm{6}{x}+\mathrm{4} \\ $$$$\mathrm{par}_{\mathrm{2}} :\:{y}=−\frac{\mathrm{13}}{\mathrm{49}}{x}^{\mathrm{2}} +\frac{\mathrm{6}}{\mathrm{49}}{x}+\frac{\mathrm{184}}{\mathrm{49}} \\ $$$$\mathrm{is}\:\mathrm{this}\:\mathrm{solution}\:\mathrm{ok},\:\mathrm{is}\:\mathrm{it}\:\mathrm{a}\:\mathrm{special}\:\mathrm{solution}\:\mathrm{in} \\ $$$$\mathrm{any}\:\mathrm{way}? \\ $$$$\mathrm{if}\:\mathrm{so},\:\mathrm{I}\:\mathrm{can}\:\mathrm{calculate}\:\mathrm{it}\:\mathrm{for}\:\mathrm{any}\:\mathrm{given}\:{a}\:\mathrm{and} \\ $$$$\mathrm{other}\:\mathrm{values}\:\mathrm{of}\:{g}\:\left(\mathrm{i}.\mathrm{e}.\:−\mathrm{9}.\mathrm{81}\right) \\ $$$$\mathrm{thank}\:\mathrm{you}! \\ $$
Commented by mr W last updated on 01/Feb/22
attempt for Q165357
$${attempt}\:{for}\:{Q}\mathrm{165357} \\ $$
Commented by mr W last updated on 03/Feb/22
Commented by mr W last updated on 03/Feb/22
note: projectile motion is reversible. the motion from start point A to end point B is the same as the motion from start point B to end point A, as if we just let the time run backward. that means when the ball starts from the ground at point B with a speed u and hits the hill at point P with a speed v, it is the same as when the ball starts from point P with speed v and hits the ground at point B with speed u. parabola y=c−(x^2 /d) collision atP (p, q) with q=c−(p^2 /d) tan θ=((2p)/d) α=(π/2)−φ+θ=ϕ+θ with ϕ=(π/2)−φ β=(π/2)−φ−θ=ϕ−θ motion from P to A: x=p−v cos α t y=q+v sin α t−((gt^2 )/2) A(0, c) p−v cos α t=0 t=(p/(v cos α)) q+v sin α−((gt^2 )/2)=c q+p tan α−((gp^2 (1+tan^2 α))/(2v^2 ))=c c−(p^2 /d)+p tan (ϕ+θ)−((gp^2 [1+tan^2 (ϕ+θ)])/(2v^2 ))=c let Φ=((2v^2 )/(gd)) ⇒1+((1+tan^2 (ϕ+θ))/Φ)−((d tan (ϕ+θ))/p)=0 ⇒Φ=((1+tan^2 (ϕ+θ))/((d/p) tan (ϕ+θ)−1)) motion from P to B: x=p+v cos β t y=q+v sin β t−((gt^2 )/2) B(b,0) with b=2a p+v cos β t=b t=((b−p)/(v cos β)) q+(b−p)tan β−((g(b−p)^2 (1+tan^2 β))/(2v^2 ))=0 ((cd−p^2 )/((b−p)^2 ))+(d/(b−p)) tan (ϕ−θ)−((1+tan^2 (ϕ−θ))/Φ)=0 ⇒Φ=((1+tan^2 (ϕ−θ))/(((cd−p^2 )/((b−p)^2 ))+(d/(b−p)) tan (ϕ−θ))) ((1+tan^2 (ϕ−θ))/(((cd−p^2 )/((b−p)^2 ))+(d/(b−p)) tan (ϕ−θ)))=((1+tan^2 (ϕ+θ))/((d/p) tan (ϕ+θ)−1)) let g=((cd−p^2 )/((b−p)^2 )), m=(d/(b−p)), n=(d/p) ((1+tan^2 (ϕ−θ))/(g+m tan (ϕ−θ)))=((1+tan^2 (ϕ+θ))/(n tan (ϕ+θ)−1)) ((1+(((tan ϕ−tan θ)/(1+tan ϕ tan θ)))^2 )/(g+m×((tan ϕ−tan θ)/(1+tan ϕ tan θ))))=((1+(((tan ϕ+tan θ)/(1−tan ϕ tan θ)))^2 )/(n×((tan ϕ+tan θ)/(1−tan ϕ tan θ))−1)) tan θ[(g+1)tan θ+m+n]tan^2 ϕ+[2(g−1)tan θ+(m−n)(1−tan^2 θ)]tan ϕ+g+1−(m+n)tan θ=0 with ξ=tan θ[(g+1)tan θ+m+n] ξ=((2p)/d)[(((cd−p^2 )/((b−p)^2 ))+1)((2p)/d)+(d/(b−p))+(d/p)] ξ=((4p^2 )/d^2 )(((cd−p^2 )/((b−p)^2 ))+1)+((2p)/(b−p))+2 ✓ η=2(g−1)tan θ+(m−n)(1−tan^2 θ) η=((4p)/d)(((cd−p^2 )/((b−p)^2 ))−1)+((d/(b−p))−(d/p))(1−((4p^2 )/d^2 )) ✓ λ=g+1−(m+n)tan θ λ=((cd−p^2 )/((b−p)^2 ))+1−((2p)/d)((d/(b−p))+(d/p)) λ=((cd−p^2 )/((b−p)^2 ))−((2p)/(b−p))−1 ✓ ⇒σ=tan ϕ=((−η+(√(η^2 −4ξλ)))/(2ξ)) Φ=((1+(((tan ϕ+tan θ)/(1−tan ϕ tan θ)))^2 )/(n×((tan ϕ+tan θ)/(1−tan ϕ tan θ))−1)) Φ=(((1−((2pσ)/d))^2 +(σ+((2p)/d))^2 )/([((d/p)+((2p)/d))σ+1](1−((2pσ)/d)))) Φ=(((1+σ^2 )(1+((4p^2 )/d^2 )))/([((d/p)+((2p)/d))σ+1](1−((2pσ)/d)))) ✓ we have u^2 =v^2 +2gq u^2 =v^2 +2g(c−(p^2 /d)) ((2u^2 )/(gd))=((2v^2 )/(gd))+(4/d)(c−(p^2 /d)) ((2u^2 )/(gd))=Φ−((4p^2 )/d^2 )+((4c)/d) (u/( (√(gd))))=U=(√((1/2)(Φ−((4p^2 )/d^2 )+((4c)/d)))) U=(√((1/2)(Φ−((4p^2 )/d^2 )+((4c)/d)))) from (dU/dp)=0 we get p corresponding to u_(min) and thus u_(min) . example: a=2m, d=1m, c=4m, b=4m U_(min) =3.0397 at p=0.3643 m u_(min) =9.612 m/s (see third diagram below)
$$\underline{{note}:} \\ $$$${projectile}\:{motion}\:{is}\:{reversible}. \\ $$$${the}\:{motion}\:{from}\:{start}\:{point}\:{A}\:{to} \\ $$$${end}\:{point}\:{B}\:{is}\:{the}\:{same}\:{as}\:{the}\:{motion} \\ $$$${from}\:{start}\:{point}\:{B}\:{to}\:{end}\:{point}\:{A},\:{as} \\ $$$${if}\:{we}\:{just}\:{let}\:{the}\:{time}\:{run}\:{backward}. \\ $$$${that}\:{means}\:{when}\:{the}\:{ball}\:{starts}\:{from} \\ $$$${the}\:{ground}\:{at}\:{point}\:{B}\:{with}\:{a}\:{speed}\:{u}\: \\ $$$${and}\:{hits}\:{the}\:{hill}\:{at}\:{point}\:{P}\:{with}\:{a}\: \\ $$$${speed}\:{v},\:{it}\:{is}\:{the}\:{same}\:{as}\:{when}\:{the}\: \\ $$$${ball}\:{starts}\:{from}\:{point}\:{P}\:{with}\:{speed}\:{v} \\ $$$${and}\:{hits}\:{the}\:{ground}\:{at}\:{point}\:{B}\:{with}\: \\ $$$${speed}\:{u}. \\ $$$${parabola}\:{y}={c}−\frac{{x}^{\mathrm{2}} }{{d}} \\ $$$${collision}\:{atP}\:\left({p},\:{q}\right)\:{with}\:{q}={c}−\frac{{p}^{\mathrm{2}} }{{d}} \\ $$$$\mathrm{tan}\:\theta=\frac{\mathrm{2}{p}}{{d}} \\ $$$$\alpha=\frac{\pi}{\mathrm{2}}−\phi+\theta=\varphi+\theta\:{with}\:\varphi=\frac{\pi}{\mathrm{2}}−\phi \\ $$$$\beta=\frac{\pi}{\mathrm{2}}−\phi−\theta=\varphi−\theta \\ $$$$ \\ $$$${motion}\:{from}\:{P}\:{to}\:{A}: \\ $$$${x}={p}−{v}\:\mathrm{cos}\:\alpha\:{t} \\ $$$${y}={q}+{v}\:\mathrm{sin}\:\alpha\:{t}−\frac{{gt}^{\mathrm{2}} }{\mathrm{2}} \\ $$$${A}\left(\mathrm{0},\:{c}\right) \\ $$$${p}−{v}\:\mathrm{cos}\:\alpha\:{t}=\mathrm{0} \\ $$$${t}=\frac{{p}}{{v}\:\mathrm{cos}\:\alpha} \\ $$$${q}+{v}\:\mathrm{sin}\:\alpha−\frac{{gt}^{\mathrm{2}} }{\mathrm{2}}={c} \\ $$$${q}+{p}\:\mathrm{tan}\:\alpha−\frac{{gp}^{\mathrm{2}} \left(\mathrm{1}+\mathrm{tan}^{\mathrm{2}} \:\alpha\right)}{\mathrm{2}{v}^{\mathrm{2}} }={c} \\ $$$${c}−\frac{{p}^{\mathrm{2}} }{{d}}+{p}\:\mathrm{tan}\:\left(\varphi+\theta\right)−\frac{{gp}^{\mathrm{2}} \left[\mathrm{1}+\mathrm{tan}^{\mathrm{2}} \:\left(\varphi+\theta\right)\right]}{\mathrm{2}{v}^{\mathrm{2}} }={c} \\ $$$${let}\:\Phi=\frac{\mathrm{2}{v}^{\mathrm{2}} }{{gd}} \\ $$$$\Rightarrow\mathrm{1}+\frac{\mathrm{1}+\mathrm{tan}^{\mathrm{2}} \:\left(\varphi+\theta\right)}{\Phi}−\frac{{d}\:\mathrm{tan}\:\left(\varphi+\theta\right)}{{p}}=\mathrm{0} \\ $$$$\Rightarrow\Phi=\frac{\mathrm{1}+\mathrm{tan}^{\mathrm{2}} \:\left(\varphi+\theta\right)}{\frac{{d}}{{p}}\:\mathrm{tan}\:\left(\varphi+\theta\right)−\mathrm{1}} \\ $$$$ \\ $$$${motion}\:{from}\:{P}\:{to}\:{B}: \\ $$$${x}={p}+{v}\:\mathrm{cos}\:\beta\:{t} \\ $$$${y}={q}+{v}\:\mathrm{sin}\:\beta\:{t}−\frac{{gt}^{\mathrm{2}} }{\mathrm{2}} \\ $$$${B}\left({b},\mathrm{0}\right)\:{with}\:{b}=\mathrm{2}{a} \\ $$$${p}+{v}\:\mathrm{cos}\:\beta\:{t}={b} \\ $$$${t}=\frac{{b}−{p}}{{v}\:\mathrm{cos}\:\beta} \\ $$$${q}+\left({b}−{p}\right)\mathrm{tan}\:\beta−\frac{{g}\left({b}−{p}\right)^{\mathrm{2}} \left(\mathrm{1}+\mathrm{tan}^{\mathrm{2}} \:\beta\right)}{\mathrm{2}{v}^{\mathrm{2}} }=\mathrm{0} \\ $$$$\frac{{cd}−{p}^{\mathrm{2}} }{\left({b}−{p}\right)^{\mathrm{2}} }+\frac{{d}}{{b}−{p}}\:\mathrm{tan}\:\left(\varphi−\theta\right)−\frac{\mathrm{1}+\mathrm{tan}^{\mathrm{2}} \:\left(\varphi−\theta\right)}{\Phi}=\mathrm{0} \\ $$$$\Rightarrow\Phi=\frac{\mathrm{1}+\mathrm{tan}^{\mathrm{2}} \:\left(\varphi−\theta\right)}{\frac{{cd}−{p}^{\mathrm{2}} }{\left({b}−{p}\right)^{\mathrm{2}} }+\frac{{d}}{{b}−{p}}\:\mathrm{tan}\:\left(\varphi−\theta\right)} \\ $$$$ \\ $$$$\frac{\mathrm{1}+\mathrm{tan}^{\mathrm{2}} \:\left(\varphi−\theta\right)}{\frac{{cd}−{p}^{\mathrm{2}} }{\left({b}−{p}\right)^{\mathrm{2}} }+\frac{{d}}{{b}−{p}}\:\mathrm{tan}\:\left(\varphi−\theta\right)}=\frac{\mathrm{1}+\mathrm{tan}^{\mathrm{2}} \:\left(\varphi+\theta\right)}{\frac{{d}}{{p}}\:\mathrm{tan}\:\left(\varphi+\theta\right)−\mathrm{1}}\:\: \\ $$$${let}\:{g}=\frac{{cd}−{p}^{\mathrm{2}} }{\left({b}−{p}\right)^{\mathrm{2}} },\:{m}=\frac{{d}}{{b}−{p}},\:{n}=\frac{{d}}{{p}} \\ $$$$\frac{\mathrm{1}+\mathrm{tan}^{\mathrm{2}} \:\left(\varphi−\theta\right)}{{g}+{m}\:\mathrm{tan}\:\left(\varphi−\theta\right)}=\frac{\mathrm{1}+\mathrm{tan}^{\mathrm{2}} \:\left(\varphi+\theta\right)}{{n}\:\mathrm{tan}\:\left(\varphi+\theta\right)−\mathrm{1}}\:\: \\ $$$$\frac{\mathrm{1}+\left(\frac{\mathrm{tan}\:\varphi−\mathrm{tan}\:\theta}{\mathrm{1}+\mathrm{tan}\:\varphi\:\mathrm{tan}\:\theta}\right)^{\mathrm{2}} }{{g}+{m}×\frac{\mathrm{tan}\:\varphi−\mathrm{tan}\:\theta}{\mathrm{1}+\mathrm{tan}\:\varphi\:\mathrm{tan}\:\theta}}=\frac{\mathrm{1}+\left(\frac{\mathrm{tan}\:\varphi+\mathrm{tan}\:\theta}{\mathrm{1}−\mathrm{tan}\:\varphi\:\mathrm{tan}\:\theta}\right)^{\mathrm{2}} }{{n}×\frac{\mathrm{tan}\:\varphi+\mathrm{tan}\:\theta}{\mathrm{1}−\mathrm{tan}\:\varphi\:\mathrm{tan}\:\theta}−\mathrm{1}}\:\: \\ $$$$\mathrm{tan}\:\theta\left[\left({g}+\mathrm{1}\right)\mathrm{tan}\:\theta+{m}+{n}\right]\mathrm{tan}^{\mathrm{2}} \:\varphi+\left[\mathrm{2}\left({g}−\mathrm{1}\right)\mathrm{tan}\:\theta+\left({m}−{n}\right)\left(\mathrm{1}−\mathrm{tan}^{\mathrm{2}} \:\theta\right)\right]\mathrm{tan}\:\varphi+{g}+\mathrm{1}−\left({m}+{n}\right)\mathrm{tan}\:\theta=\mathrm{0} \\ $$$${with} \\ $$$$\xi=\mathrm{tan}\:\theta\left[\left({g}+\mathrm{1}\right)\mathrm{tan}\:\theta+{m}+{n}\right] \\ $$$$\xi=\frac{\mathrm{2}{p}}{{d}}\left[\left(\frac{{cd}−{p}^{\mathrm{2}} }{\left({b}−{p}\right)^{\mathrm{2}} }+\mathrm{1}\right)\frac{\mathrm{2}{p}}{{d}}+\frac{{d}}{{b}−{p}}+\frac{{d}}{{p}}\right] \\ $$$$\xi=\frac{\mathrm{4}{p}^{\mathrm{2}} }{{d}^{\mathrm{2}} }\left(\frac{{cd}−{p}^{\mathrm{2}} }{\left({b}−{p}\right)^{\mathrm{2}} }+\mathrm{1}\right)+\frac{\mathrm{2}{p}}{{b}−{p}}+\mathrm{2}\:\checkmark \\ $$$$\eta=\mathrm{2}\left({g}−\mathrm{1}\right)\mathrm{tan}\:\theta+\left({m}−{n}\right)\left(\mathrm{1}−\mathrm{tan}^{\mathrm{2}} \:\theta\right) \\ $$$$\eta=\frac{\mathrm{4}{p}}{{d}}\left(\frac{{cd}−{p}^{\mathrm{2}} }{\left({b}−{p}\right)^{\mathrm{2}} }−\mathrm{1}\right)+\left(\frac{{d}}{{b}−{p}}−\frac{{d}}{{p}}\right)\left(\mathrm{1}−\frac{\mathrm{4}{p}^{\mathrm{2}} }{{d}^{\mathrm{2}} }\right)\:\checkmark \\ $$$$\lambda={g}+\mathrm{1}−\left({m}+{n}\right)\mathrm{tan}\:\theta \\ $$$$\lambda=\frac{{cd}−{p}^{\mathrm{2}} }{\left({b}−{p}\right)^{\mathrm{2}} }+\mathrm{1}−\frac{\mathrm{2}{p}}{{d}}\left(\frac{{d}}{{b}−{p}}+\frac{{d}}{{p}}\right) \\ $$$$\lambda=\frac{{cd}−{p}^{\mathrm{2}} }{\left({b}−{p}\right)^{\mathrm{2}} }−\frac{\mathrm{2}{p}}{{b}−{p}}−\mathrm{1}\:\checkmark \\ $$$$\Rightarrow\sigma=\mathrm{tan}\:\varphi=\frac{−\eta+\sqrt{\eta^{\mathrm{2}} −\mathrm{4}\xi\lambda}}{\mathrm{2}\xi} \\ $$$$\Phi=\frac{\mathrm{1}+\left(\frac{\mathrm{tan}\:\varphi+\mathrm{tan}\:\theta}{\mathrm{1}−\mathrm{tan}\:\varphi\:\mathrm{tan}\:\theta}\right)^{\mathrm{2}} }{{n}×\frac{\mathrm{tan}\:\varphi+\mathrm{tan}\:\theta}{\mathrm{1}−\mathrm{tan}\:\varphi\:\mathrm{tan}\:\theta}−\mathrm{1}} \\ $$$$\Phi=\frac{\left(\mathrm{1}−\frac{\mathrm{2}{p}\sigma}{{d}}\right)^{\mathrm{2}} +\left(\sigma+\frac{\mathrm{2}{p}}{{d}}\right)^{\mathrm{2}} }{\left[\left(\frac{{d}}{{p}}+\frac{\mathrm{2}{p}}{{d}}\right)\sigma+\mathrm{1}\right]\left(\mathrm{1}−\frac{\mathrm{2}{p}\sigma}{{d}}\right)} \\ $$$$\Phi=\frac{\left(\mathrm{1}+\sigma^{\mathrm{2}} \right)\left(\mathrm{1}+\frac{\mathrm{4}{p}^{\mathrm{2}} }{{d}^{\mathrm{2}} }\right)}{\left[\left(\frac{{d}}{{p}}+\frac{\mathrm{2}{p}}{{d}}\right)\sigma+\mathrm{1}\right]\left(\mathrm{1}−\frac{\mathrm{2}{p}\sigma}{{d}}\right)}\:\checkmark \\ $$$$ \\ $$$${we}\:{have}\: \\ $$$${u}^{\mathrm{2}} ={v}^{\mathrm{2}} +\mathrm{2}{gq} \\ $$$${u}^{\mathrm{2}} ={v}^{\mathrm{2}} +\mathrm{2}{g}\left({c}−\frac{{p}^{\mathrm{2}} }{{d}}\right) \\ $$$$\frac{\mathrm{2}{u}^{\mathrm{2}} }{{gd}}=\frac{\mathrm{2}{v}^{\mathrm{2}} }{{gd}}+\frac{\mathrm{4}}{{d}}\left({c}−\frac{{p}^{\mathrm{2}} }{{d}}\right) \\ $$$$\frac{\mathrm{2}{u}^{\mathrm{2}} }{{gd}}=\Phi−\frac{\mathrm{4}{p}^{\mathrm{2}} }{{d}^{\mathrm{2}} }+\frac{\mathrm{4}{c}}{{d}} \\ $$$$\frac{{u}}{\:\sqrt{{gd}}}={U}=\sqrt{\frac{\mathrm{1}}{\mathrm{2}}\left(\Phi−\frac{\mathrm{4}{p}^{\mathrm{2}} }{{d}^{\mathrm{2}} }+\frac{\mathrm{4}{c}}{{d}}\right)} \\ $$$${U}=\sqrt{\frac{\mathrm{1}}{\mathrm{2}}\left(\Phi−\frac{\mathrm{4}{p}^{\mathrm{2}} }{{d}^{\mathrm{2}} }+\frac{\mathrm{4}{c}}{{d}}\right)} \\ $$$${from}\:\frac{{dU}}{{dp}}=\mathrm{0}\:{we}\:{get}\:{p}\:{corresponding} \\ $$$${to}\:{u}_{{min}} \:{and}\:{thus}\:{u}_{{min}} . \\ $$$$ \\ $$$$\underline{{example}:}\: \\ $$$${a}=\mathrm{2}{m},\:{d}=\mathrm{1}{m},\:{c}=\mathrm{4}{m},\:{b}=\mathrm{4}{m} \\ $$$${U}_{{min}} =\mathrm{3}.\mathrm{0397}\:{at}\:{p}=\mathrm{0}.\mathrm{3643}\:{m} \\ $$$${u}_{{min}} =\mathrm{9}.\mathrm{612}\:{m}/{s} \\ $$$$\left({see}\:{third}\:{diagram}\:{below}\right) \\ $$
Commented by mr W last updated on 02/Feb/22
Commented by mr W last updated on 02/Feb/22
Commented by mr W last updated on 03/Feb/22
Commented by MJS_new last updated on 02/Feb/22
am I wrong here? both green parabolas must have the same second derivate which is y_1 ^(′′) =y_2 ′′=−9.81. under this condition (and the other given conditions) there′s no solution.
$$\mathrm{am}\:\mathrm{I}\:\mathrm{wrong}\:\mathrm{here}?\:\mathrm{both}\:\mathrm{green}\:\mathrm{parabolas}\:\mathrm{must} \\ $$$$\mathrm{have}\:\mathrm{the}\:\mathrm{same}\:\mathrm{second}\:\mathrm{derivate}\:\mathrm{which}\:\mathrm{is} \\ $$$${y}_{\mathrm{1}} ^{''} ={y}_{\mathrm{2}} ''=−\mathrm{9}.\mathrm{81}.\:\mathrm{under}\:\mathrm{this}\:\mathrm{condition}\:\left(\mathrm{and}\right. \\ $$$$\left.\mathrm{the}\:\mathrm{other}\:\mathrm{given}\:\mathrm{conditions}\right)\:\mathrm{there}'\mathrm{s}\:\mathrm{no}\:\mathrm{solution}. \\ $$
Commented by mr W last updated on 02/Feb/22
it′s not correct with y′′=−g sir. before and after point P the ball has the same speed, but different direction angles to horizontal, therefore both green parabolas even must have different y′′ at this point. the diagrams show the real path of the ball when started with different initial speed. actually there are infinite many possible pathes for the ball, in one of them the ball needs the minimum energy to reach the peak of hill. approximately the highest point of green parabola shows the energy of the ball. so in the third path the ball needs less energy than in the other pathes.
$${it}'{s}\:{not}\:{correct}\:{with}\:{y}''=−{g}\:{sir}. \\ $$$${before}\:{and}\:{after}\:{point}\:{P}\:\:{the}\:{ball}\:{has} \\ $$$${the}\:{same}\:{speed},\:{but}\:{different}\: \\ $$$${direction}\:{angles}\:{to}\:{horizontal}, \\ $$$${therefore}\:{both}\:{green}\:{parabolas}\:{even} \\ $$$${must}\:{have}\:{different}\:{y}''\:{at}\:{this}\:{point}. \\ $$$${the}\:{diagrams}\:{show}\:{the}\:{real}\:{path}\:{of} \\ $$$${the}\:{ball}\:{when}\:{started}\:{with}\:{different} \\ $$$${initial}\:{speed}.\:{actually}\:{there}\:{are}\:{infinite} \\ $$$${many}\:{possible}\:{pathes}\:{for}\:{the}\:{ball}, \\ $$$${in}\:{one}\:{of}\:{them}\:{the}\:{ball}\:{needs}\:{the} \\ $$$${minimum}\:{energy}\:{to}\:{reach}\:{the}\:{peak} \\ $$$${of}\:{hill}. \\ $$$${approximately}\:{the}\:{highest}\:{point}\:{of}\: \\ $$$${green}\:{parabola}\:{shows}\:{the}\:{energy}\:{of}\: \\ $$$${the}\:{ball}.\:{so}\:{in}\:{the}\:{third}\:{path}\:{the}\:{ball}\: \\ $$$${needs}\:{less}\:{energy}\:{than}\:{in}\:{the}\:{other} \\ $$$${pathes}. \\ $$
Commented by MJS_new last updated on 02/Feb/22
but the only acceleration is g. pkease explain this to me: if the parabola is s then v should be s′ and a=−g should be s′′ I′m not good in these things...
$$\mathrm{but}\:\mathrm{the}\:\mathrm{only}\:\mathrm{acceleration}\:\mathrm{is}\:{g}.\:\mathrm{pkease} \\ $$$$\mathrm{explain}\:\mathrm{this}\:\mathrm{to}\:\mathrm{me}:\:\mathrm{if}\:\mathrm{the}\:\mathrm{parabola}\:\mathrm{is}\:{s} \\ $$$$\mathrm{then}\:{v}\:\mathrm{should}\:\mathrm{be}\:{s}'\:\mathrm{and}\:{a}=−{g}\:\mathrm{should}\:\mathrm{be}\:{s}'' \\ $$$$\mathrm{I}'\mathrm{m}\:\mathrm{not}\:\mathrm{good}\:\mathrm{in}\:\mathrm{these}\:\mathrm{things}… \\ $$
Commented by MJS_new last updated on 02/Feb/22
...haha, forgot that y′′ also depends on v_0 in x−direction. as I mentioned, I′m not good at physics
$$…\mathrm{haha},\:\mathrm{forgot}\:\mathrm{that}\:{y}''\:\mathrm{also}\:\mathrm{depends}\:\mathrm{on}\:{v}_{\mathrm{0}} \\ $$$$\mathrm{in}\:{x}−\mathrm{direction}.\:\mathrm{as}\:\mathrm{I}\:\mathrm{mentioned},\:\mathrm{I}'\mathrm{m}\:\mathrm{not} \\ $$$$\mathrm{good}\:\mathrm{at}\:\mathrm{physics} \\ $$
Commented by Tawa11 last updated on 02/Feb/22
Great sir.
$$\mathrm{Great}\:\mathrm{sir}. \\ $$
Commented by mr W last updated on 02/Feb/22
MJS sir: exactly! when the ball hits an inclined surface, the horizontal component of speed changes, therefore y_1 ′′≠y_2 ′′. the horizontal component of speed decides if the parabola is flat or steep, i.e. if y′′ is small or large.
$${MJS}\:{sir}: \\ $$$${exactly}! \\ $$$${when}\:{the}\:{ball}\:{hits}\:{an}\:{inclined}\: \\ $$$${surface},\:{the}\:{horizontal}\:{component} \\ $$$${of}\:{speed}\:{changes},\:{therefore}\:{y}_{\mathrm{1}} ''\neq{y}_{\mathrm{2}} ''. \\ $$$${the}\:{horizontal}\:{component}\:{of}\:{speed} \\ $$$${decides}\:{if}\:{the}\:{parabola}\:{is}\:{flat}\:{or}\:{steep}, \\ $$$${i}.{e}.\:{if}\:{y}''\:{is}\:{small}\:{or}\:{large}. \\ $$
Commented by mr W last updated on 02/Feb/22
Commented by mr W last updated on 02/Feb/22
here e stands for restitution coef. of a collision. “e=1” means that the collision is perfectly elastic, no energy goes lost. the ball behaves like a light ray.
$${here}\:{e}\:{stands}\:{for}\:{restitution}\:{coef}.\:{of} \\ $$$${a}\:{collision}. \\ $$$$“{e}=\mathrm{1}''\:{means}\:{that}\:{the}\:{collision}\:{is} \\ $$$${perfectly}\:{elastic},\:{no}\:{energy}\:{goes}\:{lost}. \\ $$$${the}\:{ball}\:{behaves}\:{like}\:{a}\:{light}\:{ray}. \\ $$
Commented by MJS_new last updated on 02/Feb/22
ok. anyway my solution should hold, because I assumed an elastic collision. but does my solution match yours in this special case?
$$\mathrm{ok}.\:\mathrm{anyway}\:\mathrm{my}\:\mathrm{solution}\:\mathrm{should}\:\mathrm{hold},\:\mathrm{because} \\ $$$$\mathrm{I}\:\mathrm{assumed}\:\mathrm{an}\:\mathrm{elastic}\:\mathrm{collision}.\:\mathrm{but}\:\mathrm{does}\:\mathrm{my} \\ $$$$\mathrm{solution}\:\mathrm{match}\:\mathrm{yours}\:\mathrm{in}\:\mathrm{this}\:\mathrm{special}\:\mathrm{case}? \\ $$
Commented by mr W last updated on 02/Feb/22
yes, your solution is a possible path of the ball.but it is not the path of the ball with the minimum initial speed.
$${yes},\:{your}\:{solution}\:{is}\:{a}\:{possible}\:{path} \\ $$$${of}\:{the}\:{ball}.{but}\:{it}\:{is}\:{not}\:{the}\:{path}\:{of} \\ $$$${the}\:{ball}\:{with}\:{the}\:{minimum}\:{initial} \\ $$$${speed}. \\ $$
Commented by MJS_new last updated on 02/Feb/22
ok. is it the only path with reflection at x=(1/2) (with g=−10 and a=2)?
$$\mathrm{ok}.\:\mathrm{is}\:\mathrm{it}\:\mathrm{the}\:\mathrm{only}\:\mathrm{path}\:\mathrm{with}\:\mathrm{reflection}\:\mathrm{at} \\ $$$${x}=\frac{\mathrm{1}}{\mathrm{2}}\:\left(\mathrm{with}\:{g}=−\mathrm{10}\:\mathrm{and}\:{a}=\mathrm{2}\right)? \\ $$
Commented by mr W last updated on 02/Feb/22
yes!
$${yes}! \\ $$
Answered by mr W last updated on 03/Feb/22
Summary of solution parabola y=c−(x^2 /d) point on top of parabola A(0, c) point on ground B(b, 0) initial speed at point B is u. collision point P at x=p 0<p<((4c−d+(√((4c−d)^2 +(4bd)^2 )))/(8b)) let ξ=((4p^2 )/d^2 )(((cd−p^2 )/((b−p)^2 ))+1)+((2p)/(b−p))+2 η=((4p)/d)(((cd−p^2 )/((b−p)^2 ))−1)+((d/(b−p))−(d/p))(1−((4p^2 )/d^2 )) λ=((cd−p^2 )/((b−p)^2 ))−((2p)/(b−p))−1 σ=((−η+(√(η^2 −4ξλ)))/(2ξ)) Φ=(((1+σ^2 )(1+((4p^2 )/d^2 )))/([((d/p)+((2p)/d))σ+1](1−((2pσ)/d)))) U=(u/( (√(gd))))=(√((1/2)(Φ−((4p^2 )/d^2 )+((4c)/d)))) U is a function of parameter p. it has a minimun. from (dU/dp)=0 we find U_(min) and thus u_(min) . see following example:
$$\underline{{Summary}\:{of}\:{solution}} \\ $$$${parabola}\:{y}={c}−\frac{{x}^{\mathrm{2}} }{{d}} \\ $$$${point}\:{on}\:{top}\:{of}\:{parabola}\:{A}\left(\mathrm{0},\:{c}\right) \\ $$$${point}\:{on}\:{ground}\:{B}\left({b},\:\mathrm{0}\right) \\ $$$${initial}\:{speed}\:{at}\:{point}\:{B}\:{is}\:{u}. \\ $$$${collision}\:{point}\:{P}\:{at}\:{x}={p} \\ $$$$\mathrm{0}<{p}<\frac{\mathrm{4}{c}−{d}+\sqrt{\left(\mathrm{4}{c}−{d}\right)^{\mathrm{2}} +\left(\mathrm{4}{bd}\right)^{\mathrm{2}} }}{\mathrm{8}{b}} \\ $$$${let} \\ $$$$\xi=\frac{\mathrm{4}{p}^{\mathrm{2}} }{{d}^{\mathrm{2}} }\left(\frac{{cd}−{p}^{\mathrm{2}} }{\left({b}−{p}\right)^{\mathrm{2}} }+\mathrm{1}\right)+\frac{\mathrm{2}{p}}{{b}−{p}}+\mathrm{2} \\ $$$$\eta=\frac{\mathrm{4}{p}}{{d}}\left(\frac{{cd}−{p}^{\mathrm{2}} }{\left({b}−{p}\right)^{\mathrm{2}} }−\mathrm{1}\right)+\left(\frac{{d}}{{b}−{p}}−\frac{{d}}{{p}}\right)\left(\mathrm{1}−\frac{\mathrm{4}{p}^{\mathrm{2}} }{{d}^{\mathrm{2}} }\right) \\ $$$$\lambda=\frac{{cd}−{p}^{\mathrm{2}} }{\left({b}−{p}\right)^{\mathrm{2}} }−\frac{\mathrm{2}{p}}{{b}−{p}}−\mathrm{1} \\ $$$$\sigma=\frac{−\eta+\sqrt{\eta^{\mathrm{2}} −\mathrm{4}\xi\lambda}}{\mathrm{2}\xi} \\ $$$$\Phi=\frac{\left(\mathrm{1}+\sigma^{\mathrm{2}} \right)\left(\mathrm{1}+\frac{\mathrm{4}{p}^{\mathrm{2}} }{{d}^{\mathrm{2}} }\right)}{\left[\left(\frac{{d}}{{p}}+\frac{\mathrm{2}{p}}{{d}}\right)\sigma+\mathrm{1}\right]\left(\mathrm{1}−\frac{\mathrm{2}{p}\sigma}{{d}}\right)} \\ $$$${U}=\frac{{u}}{\:\sqrt{{gd}}}=\sqrt{\frac{\mathrm{1}}{\mathrm{2}}\left(\Phi−\frac{\mathrm{4}{p}^{\mathrm{2}} }{{d}^{\mathrm{2}} }+\frac{\mathrm{4}{c}}{{d}}\right)} \\ $$$${U}\:{is}\:{a}\:{function}\:{of}\:{parameter}\:{p}.\:{it}\:{has} \\ $$$${a}\:{minimun}. \\ $$$${from}\:\frac{{dU}}{{dp}}=\mathrm{0}\:{we}\:{find}\:{U}_{{min}} \:{and}\:{thus}\:{u}_{{min}} . \\ $$$${see}\:{following}\:{example}: \\ $$
Commented by mr W last updated on 03/Feb/22
Commented by mr W last updated on 03/Feb/22
Commented by ajfour last updated on 03/Feb/22
I′ll go through all this sir, given any dynamics, there is a transport lag...<_/ (⌣_( (√)) ^(• •) )>_\
$${I}'{ll}\:{go}\:{through}\:{all}\:{this}\:{sir},\:{given} \\ $$$${any}\:{dynamics},\:{there}\:{is}\:{a}\:{transport} \\ $$$${lag}…\underset{/} {<}\left(\underset{\:\sqrt{}} {\overset{\bullet\:\:\:\bullet} {\smile}}\right)\underset{\backslash} {>} \\ $$$$ \\ $$
Commented by mr W last updated on 03/Feb/22
thanks for attention sir! this is the only way i found to express the speed u in terms of only one variable.
$${thanks}\:{for}\:{attention}\:{sir}! \\ $$$${this}\:{is}\:{the}\:{only}\:{way}\:{i}\:{found}\:{to}\: \\ $$$${express}\:{the}\:{speed}\:{u}\:{in}\:{terms}\:{of}\:{only} \\ $$$${one}\:{variable}. \\ $$

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