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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

$$\mathrm{I}\mathrm{would}\mathrm{like}\mathrm{you}\mathrm{to}\mathrm{check}\mathrm{this}:$$$$\mathrm{from}\mathrm{Sir}{\mathrm{Ajfour}}^{\prime }\mathrm{s}\mathrm{scetch}\mathrm{I}\mathrm{assumed}\mathrm{that}\mathrm{the}$$$$‘‘{\mathrm{hill}}^{″}\mathrm{is}y=-x^2+a^2\mathrm{and}\mathrm{the}\mathrm{intersection}\mathrm{is}$$$$\mathrm{at}x=\frac{1}{2}\mathrm{because}\mathrm{of}\mathrm{the}‘‘e=1^{″}-\mathrm{mark};\mathrm{I}$$$$\mathrm{understood}\mathrm{that}\mathrm{the}\mathrm{reflecting}\mathrm{tangent}\mathrm{has}$$$$\mathrm{slope}-1$$$$\mathrm{from}\mathrm{there}\mathrm{I}\mathrm{found}\mathrm{this}:$$$$\mathrm{for}g=-10\mathrm{and}a=2\mathrm{I}\mathrm{get}\mathrm{these}\mathrm{formulas}:$$$$‘‘{\mathrm{hill}}^{″}:y=-x^2+4$$$${\mathrm{par}}_1:y=-13x^2+6x+4$$$${\mathrm{par}}_2:y=-\frac{13}{49}{x}^2+\frac{6}{49}x+\frac{184}{49}$$$$\mathrm{is}\mathrm{this}\mathrm{solution}\mathrm{ok},\mathrm{is}\mathrm{it}\mathrm{a}\mathrm{special}\mathrm{solution}\mathrm{in}$$$$\mathrm{any}\mathrm{way}?$$$$\mathrm{if}\mathrm{so},\mathrm{I}\mathrm{can}\mathrm{calculate}\mathrm{it}\mathrm{for}\mathrm{any}\mathrm{given}a\mathrm{and}$$$$\mathrm{other}\mathrm{values}\mathrm{of}g(\mathrm{i}.\mathrm{e}.-9.81)$$$$\mathrm{thank}\mathrm{you}!$$

Commented by mr W last updated on 01/Feb/22

$$attemptforQ165357$$

Commented by mr W last updated on 03/Feb/22

Commented by mr W last updated on 03/Feb/22

$$\underset{―}{note:}$$$$projectilemotionisreversible.$$$$themotionfromstartpointAto$$$$endpointBisthesameasthemotion$$$$fromstartpointBtoendpointA,as$$$$ifwejustletthetimerunbackward.$$$$thatmeanswhentheballstartsfrom$$$$thegroundatpointBwithaspeedu$$$$andhitsthehillatpointPwitha$$$$speedv,itisthesameaswhenthe$$$$ballstartsfrompointPwithspeedv$$$$andhitsthegroundatpointBwith$$$$speedu.$$$$parabolay=c-\frac{x^2}{d}$$$$collisionatP(p,q)withq=c-\frac{p^2}{d}$$$$\tan \theta =\frac{2p}{d}$$$$\alpha =\frac{\pi }{2}-\varphi +\theta =\phi +\theta with\phi =\frac{\pi }{2}-\varphi$$$$\beta =\frac{\pi }{2}-\varphi -\theta =\phi -\theta$$$$$$$$motionfromPtoA:$$$$x=p-v\cos \alpha t$$$$y=q+v\sin \alpha t-\frac{{gt}^2}{2}$$$$A(0,c)$$$$p-v\cos \alpha t=0$$$$t=\frac{p}{v\cos \alpha }$$$$q+v\sin \alpha -\frac{{gt}^2}{2}=c$$$$q+p\tan \alpha -\frac{{gp}^2(1+{\tan }^2\alpha )}{2v^2}=c$$$$c-\frac{p^2}{d}+p\tan (\phi +\theta )-\frac{{gp}^2[1+{\tan }^2(\phi +\theta )]}{2v^2}=c$$$$let\mathrm{\Phi }=\frac{2v^2}{gd}$$$$\Rightarrow 1+\frac{1+{\tan }^2(\phi +\theta )}{\mathrm{\Phi }}-\frac{d\tan (\phi +\theta )}{p}=0$$$$\Rightarrow \mathrm{\Phi }=\frac{1+{\tan }^2(\phi +\theta )}{\frac{d}{p}\tan (\phi +\theta )-1}$$$$$$$$motionfromPtoB:$$$$x=p+v\cos \beta t$$$$y=q+v\sin \beta t-\frac{{gt}^2}{2}$$$$B(b,0)withb=2a$$$$p+v\cos \beta t=b$$$$t=\frac{b-p}{v\cos \beta }$$$$q+(b-p)\tan \beta -\frac{g{(b-p)}^2(1+{\tan }^2\beta )}{2v^2}=0$$$$\frac{cd-p^2}{{(b-p)}^2}+\frac{d}{b-p}\tan (\phi -\theta )-\frac{1+{\tan }^2(\phi -\theta )}{\mathrm{\Phi }}=0$$$$\Rightarrow \mathrm{\Phi }=\frac{1+{\tan }^2(\phi -\theta )}{\frac{cd-p^2}{{(b-p)}^2}+\frac{d}{b-p}\tan (\phi -\theta )}$$$$$$$$\frac{1+{\tan }^2(\phi -\theta )}{\frac{cd-p^2}{{(b-p)}^2}+\frac{d}{b-p}\tan (\phi -\theta )}=\frac{1+{\tan }^2(\phi +\theta )}{\frac{d}{p}\tan (\phi +\theta )-1}$$$$letg=\frac{cd-p^2}{{(b-p)}^2},m=\frac{d}{b-p},n=\frac{d}{p}$$$$\frac{1+{\tan }^2(\phi -\theta )}{g+m\tan (\phi -\theta )}=\frac{1+{\tan }^2(\phi +\theta )}{n\tan (\phi +\theta )-1}$$$$\frac{1+{\left(\frac{\tan \phi -\tan \theta }{1+\tan \phi \tan \theta }\right)}^2}{g+m\times \frac{\tan \phi -\tan \theta }{1+\tan \phi \tan \theta }}=\frac{1+{\left(\frac{\tan \phi +\tan \theta }{1-\tan \phi \tan \theta }\right)}^2}{n\times \frac{\tan \phi +\tan \theta }{1-\tan \phi \tan \theta }-1}$$$$\tan \theta [(g+1)\tan \theta +m+n]{\tan }^2\phi +[2(g-1)\tan \theta +(m-n)(1-{\tan }^2\theta )]\tan \phi +g+1-(m+n)\tan \theta =0$$$$with$$$$\xi =\tan \theta [(g+1)\tan \theta +m+n]$$$$\xi =\frac{2p}{d}[(\frac{cd-p^2}{{(b-p)}^2}+1)\frac{2p}{d}+\frac{d}{b-p}+\frac{d}{p}]$$$$\xi =\frac{4p^2}{d^2}(\frac{cd-p^2}{{(b-p)}^2}+1)+\frac{2p}{b-p}+2✓$$$$\eta =2(g-1)\tan \theta +(m-n)(1-{\tan }^2\theta )$$$$\eta =\frac{4p}{d}(\frac{cd-p^2}{{(b-p)}^2}-1)+(\frac{d}{b-p}-\frac{d}{p})(1-\frac{4p^2}{d^2})✓$$$$\lambda =g+1-(m+n)\tan \theta$$$$\lambda =\frac{cd-p^2}{{(b-p)}^2}+1-\frac{2p}{d}(\frac{d}{b-p}+\frac{d}{p})$$$$\lambda =\frac{cd-p^2}{{(b-p)}^2}-\frac{2p}{b-p}-1✓$$$$\Rightarrow \sigma =\tan \phi =\frac{-\eta +\sqrt{{\eta }^2-4\xi \lambda }}{2\xi }$$$$\mathrm{\Phi }=\frac{1+{\left(\frac{\tan \phi +\tan \theta }{1-\tan \phi \tan \theta }\right)}^2}{n\times \frac{\tan \phi +\tan \theta }{1-\tan \phi \tan \theta }-1}$$$$\mathrm{\Phi }=\frac{{(1-\frac{2p\sigma }{d})}^2+{(\sigma +\frac{2p}{d})}^2}{[(\frac{d}{p}+\frac{2p}{d})\sigma +1](1-\frac{2p\sigma }{d})}$$$$\mathrm{\Phi }=\frac{(1+{\sigma }^2)(1+\frac{4p^2}{d^2})}{[(\frac{d}{p}+\frac{2p}{d})\sigma +1](1-\frac{2p\sigma }{d})}✓$$$$$$$$wehave$$$$u^2=v^2+2gq$$$$u^2=v^2+2g(c-\frac{p^2}{d})$$$$\frac{2u^2}{gd}=\frac{2v^2}{gd}+\frac{4}{d}(c-\frac{p^2}{d})$$$$\frac{2u^2}{gd}=\mathrm{\Phi }-\frac{4p^2}{d^2}+\frac{4c}{d}$$$$\frac{u}{\sqrt{gd}}=U=\sqrt{\frac{1}{2}(\mathrm{\Phi }-\frac{4p^2}{d^2}+\frac{4c}{d})}$$$$U=\sqrt{\frac{1}{2}(\mathrm{\Phi }-\frac{4p^2}{d^2}+\frac{4c}{d})}$$$$from\frac{dU}{dp}=0wegetpcorresponding$$$$to{u}_{min}andthus{u}_{min}.$$$$$$$$\underset{―}{example:}$$$$a=2m,d=1m,c=4m,b=4m$$$$U_{min}=3.0397atp=0.3643m$$$$u_{min}=9.612m/s$$$$\left(seethirddiagrambelow\right)$$

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

$$\mathrm{am}\mathrm{I}\mathrm{wrong}\mathrm{here}?\mathrm{both}\mathrm{green}\mathrm{parabolas}\mathrm{must}$$$$\mathrm{have}\mathrm{the}\mathrm{same}\mathrm{second}\mathrm{derivate}\mathrm{which}\mathrm{is}$$$$y_1^{{}^{″}}=y_2^{″}=-9.81.\mathrm{under}\mathrm{this}\mathrm{condition}(\mathrm{and}$$$$\mathrm{the}\mathrm{other}\mathrm{given}\mathrm{conditions}){\mathrm{there}}^{\prime }\mathrm{s}\mathrm{no}\mathrm{solution}.$$

Commented by mr W last updated on 02/Feb/22

$${it}^{\prime }snotcorrectwith{y}^{″}=-gsir.$$$$beforeandafterpointPtheballhas$$$$thesamespeed,butdifferent$$$$directionanglestohorizontal,$$$$thereforebothgreenparabolaseven$$$$musthavedifferent{y}^{″}atthispoint.$$$$thediagramsshowtherealpathof$$$$theballwhenstartedwithdifferent$$$$initialspeed.actuallythereareinfinite$$$$manypossiblepathesfortheball,$$$$inoneofthemtheballneedsthe$$$$minimumenergytoreachthepeak$$$$ofhill.$$$$approximatelythehighestpointof$$$$greenparabolashowstheenergyof$$$$theball.sointhethirdpaththeball$$$$needslessenergythanintheother$$$$pathes.$$

Commented by MJS_new last updated on 02/Feb/22

$$\mathrm{but}\mathrm{the}\mathrm{only}\mathrm{acceleration}\mathrm{is}g.\mathrm{pkease}$$$$\mathrm{explain}\mathrm{this}\mathrm{to}\mathrm{me}:\mathrm{if}\mathrm{the}\mathrm{parabola}\mathrm{is}s$$$$\mathrm{then}v\mathrm{should}\mathrm{be}{s}^{\prime }\mathrm{and}a=-g\mathrm{should}\mathrm{be}{s}^{″}$$$${\mathrm{I}}^{\prime }\mathrm{m}\mathrm{not}\mathrm{good}\mathrm{in}\mathrm{these}\mathrm{things}...$$

Commented by MJS_new last updated on 02/Feb/22

$$...\mathrm{haha},\mathrm{forgot}\mathrm{that}{y}^{″}\mathrm{also}\mathrm{depends}\mathrm{on}{v}_0$$$$\mathrm{in}x-\mathrm{direction}.\mathrm{as}\mathrm{I}\mathrm{mentioned},{\mathrm{I}}^{\prime }\mathrm{m}\mathrm{not}$$$$\mathrm{good}\mathrm{at}\mathrm{physics}$$

Commented by Tawa11 last updated on 02/Feb/22

$$\mathrm{Great}\mathrm{sir}.$$

Commented by mr W last updated on 02/Feb/22

$$MJSsir:$$$$exactly!$$$$whentheballhitsaninclined$$$$surface,thehorizontalcomponent$$$$ofspeedchanges,therefore{y}_1^{″}\ne y_2^{″}.$$$$thehorizontalcomponentofspeed$$$$decidesiftheparabolaisflatorsteep,$$$$i.e.if{y}^{″}issmallorlarge.$$

Commented by mr W last updated on 02/Feb/22

Commented by mr W last updated on 02/Feb/22

$$hereestandsforrestitutioncoef.of$$$$acollision.$$$$‘‘e=1^{″}meansthatthecollisionis$$$$perfectlyelastic,noenergygoeslost.$$$$theballbehaveslikealightray.$$

Commented by MJS_new last updated on 02/Feb/22

$$\mathrm{ok}.\mathrm{anyway}\mathrm{my}\mathrm{solution}\mathrm{should}\mathrm{hold},\mathrm{because}$$$$\mathrm{I}\mathrm{assumed}\mathrm{an}\mathrm{elastic}\mathrm{collision}.\mathrm{but}\mathrm{does}\mathrm{my}$$$$\mathrm{solution}\mathrm{match}\mathrm{yours}\mathrm{in}\mathrm{this}\mathrm{special}\mathrm{case}?$$

Commented by mr W last updated on 02/Feb/22

$$yes,yoursolutionisapossiblepath$$$$oftheball.butitisnotthepathof$$$$theballwiththeminimuminitial$$$$speed.$$

Commented by MJS_new last updated on 02/Feb/22

$$\mathrm{ok}.\mathrm{is}\mathrm{it}\mathrm{the}\mathrm{only}\mathrm{path}\mathrm{with}\mathrm{reflection}\mathrm{at}$$$$x=\frac{1}{2}(\mathrm{with}g=-10\mathrm{and}a=2)?$$

Commented by mr W last updated on 02/Feb/22

$$yes!$$

Answered by mr W last updated on 03/Feb/22

$$\underset{―}{Summaryofsolution}$$$$parabolay=c-\frac{x^2}{d}$$$$pointontopofparabolaA(0,c)$$$$pointongroundB(b,0)$$$$initialspeedatpointBisu.$$$$collisionpointPatx=p$$$$0<p<\frac{4c-d+\sqrt{{(4c-d)}^2+{\left(4bd\right)}^2}}{8b}$$$$let$$$$\xi =\frac{4p^2}{d^2}(\frac{cd-p^2}{{(b-p)}^2}+1)+\frac{2p}{b-p}+2$$$$\eta =\frac{4p}{d}(\frac{cd-p^2}{{(b-p)}^2}-1)+(\frac{d}{b-p}-\frac{d}{p})(1-\frac{4p^2}{d^2})$$$$\lambda =\frac{cd-p^2}{{(b-p)}^2}-\frac{2p}{b-p}-1$$$$\sigma =\frac{-\eta +\sqrt{{\eta }^2-4\xi \lambda }}{2\xi }$$$$\mathrm{\Phi }=\frac{(1+{\sigma }^2)(1+\frac{4p^2}{d^2})}{[(\frac{d}{p}+\frac{2p}{d})\sigma +1](1-\frac{2p\sigma }{d})}$$$$U=\frac{u}{\sqrt{gd}}=\sqrt{\frac{1}{2}(\mathrm{\Phi }-\frac{4p^2}{d^2}+\frac{4c}{d})}$$$$Uisafunctionofparameterp.ithas$$$$aminimun.$$$$from\frac{dU}{dp}=0wefind{U}_{min}andthus{u}_{min}.$$$$seefollowingexample:$$

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

$$I^{\prime }llgothroughallthissir,given$$$$anydynamics,thereisatransport$$$$lag...\underset{/}{<}\left(\underset{\sqrt{}}{\stackrel{\bullet \bullet }{⌣}}\right)\underset{\mathrm{\setminus }}{>}$$$$$$

Commented by mr W last updated on 03/Feb/22

$$thanksforattentionsir!$$$$thisistheonlywayifoundto$$$$expressthespeeduintermsofonly$$$$onevariable.$$

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