Menu Close

Question-165457




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.
$${thanks}\:{for}\:{attention}\:{sir}! \\ $$$${this}\:{is}\:{the}\:{only}\:{way}\:{i}\:{found}\:{to}\: \\ $$$${express}\:{the}\:{speed}\:{u}\:{in}\:{terms}\:{of}\:{only} \\ $$$${one}\:{variable}. \\ $$

Leave a Reply

Your email address will not be published. Required fields are marked *