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Question Number 65589 by mr W last updated on 31/Jul/19

Commented by mr W last updated on 31/Jul/19

$$AnswertoQ65277$$

Commented by Tanmay chaudhury last updated on 31/Jul/19

$$sirplsexplainifpossible..$$

Commented by mr W last updated on 12/Feb/21

$$sincethecollisioniselastic,the$$$$velocityoftheballafterthecollision$$$$isthesameasbeforethecollision.$$$$thatmeanstheballreturnstoits$$$$startpositionfollowingthesame$$$$traceasitcame.thatmeansalso$$$$thatthedirectionofthevelocityatthe$$$$pointofcollisionmustbeperpendicular$$$$tothesurface.$$$$$$$$lett=timefromAtoP$$$$t=\frac{R(1+\cos \alpha )}{u\cos \theta }$$$$atpointP:$$$$v_x=u\cos \theta$$$$v_y=-u\sin \theta +\frac{gR(1+\cos \alpha )}{u\cos \theta }$$$$v_y=v_x\tan \alpha$$$$\Rightarrow -u\sin \theta +\frac{gR(1+\cos \alpha )}{u\cos \theta }=u\cos \theta \tan \alpha$$$$\Rightarrow \frac{gR(1+\cos \alpha )}{u^2\cos \theta }=\sin \theta +\cos \theta \tan \alpha$$$$with\lambda =\frac{gR}{u^2}$$$$\Rightarrow \frac{\lambda (1+\cos \alpha )}{\cos \theta }=\sin \theta +\cos \theta \tan \alpha$$$$\Rightarrow \lambda =\frac{\cos \theta (\sin \theta +\cos \theta \tan \alpha )}{1+\cos \alpha }$$$$\Rightarrow \frac{u^2}{gR}=\frac{1+\cos \alpha }{(\tan \theta +\tan \alpha ){\cos }^2\theta }...\left(i\right)$$$$$$$$-R\sin \alpha =u\sin \theta t-\frac{1}{2}{gt}^2$$$$\Rightarrow t=\frac{u\sin \theta +\sqrt{u^2{\sin }^2\theta +2gR\sin \alpha }}{g}$$$$\Rightarrow \frac{u\sin \theta +\sqrt{u^2{\sin }^2\theta +2gR\sin \alpha }}{g}=\frac{R(1+\cos \alpha )}{u\cos \theta }$$$$\Rightarrow \sin \theta +\sqrt{{\sin }^2\theta +2\frac{gR}{u^2}\sin \alpha }=\frac{gR}{u^2}\times \frac{(1+\cos \alpha )}{\cos \theta }$$$$\Rightarrow \sin \theta +\sqrt{{\sin }^2\theta +2\lambda \sin \alpha }=\frac{\lambda (1+\cos \alpha )}{\cos \theta }$$$$\Rightarrow \sin \theta +\sqrt{{\sin }^2\theta +2\lambda \sin \alpha }=\sin \theta +\cos \theta \tan \alpha$$$$\Rightarrow \sqrt{{\sin }^2\theta +2\lambda \sin \alpha }=\cos \theta \tan \alpha$$$$\Rightarrow {\sin }^2\theta +2\lambda \sin \alpha ={\left(\cos \theta \tan \alpha \right)}^2$$$$\Rightarrow \lambda =\frac{{\left(\cos \theta \tan \alpha \right)}^2-{\sin }^2\theta }{2\sin \alpha }$$$$\Rightarrow \frac{{\left(\cos \theta \tan \alpha \right)}^2-{\sin }^2\theta }{2\sin \alpha }=\frac{\cos \theta (\sin \theta +\cos \theta \tan \alpha )}{1+\cos \alpha }$$$$\Rightarrow \frac{\cos \theta \tan \alpha -\sin \theta }{2\sin \alpha }=\frac{\cos \theta }{1+\cos \alpha }$$$$\Rightarrow (\tan \alpha -\sin \alpha )\cos \theta -(1+\cos \alpha )\sin \theta =0$$$$\Rightarrow \tan \theta =\frac{\tan \alpha -\sin \alpha }{1+\cos \alpha }...\left(ii\right)$$$$putthisinto\left(i\right):$$$$\frac{u^2}{gR}=\frac{(1+\cos \alpha )(1+{\tan }^2\theta )}{(\tan \alpha +\tan \theta )}$$$$\Rightarrow \frac{u^2}{gR}=\frac{{(1+\cos \alpha )}^2+{(\tan \alpha -\sin \alpha )}^2}{2\tan \alpha }$$$$minimumofuisat:$$$$\alpha =64.2057°$$$$\theta =39.159°$$$$u_{min}=0.9098\sqrt{gR}$$

Commented by Tanmay chaudhury last updated on 31/Jul/19

$$thankyousir...thisproblemremainunsolved$$$$tilldate...youdiditexcellent...$$

Commented by mr W last updated on 01/Aug/19

Commented by mr W last updated on 01/Aug/19

Commented by mr W last updated on 01/Aug/19

Commented by mr W last updated on 01/Aug/19

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