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Question Number 53811 by ajfour last updated on 26/Jan/19
Commented by ajfour last updated on 26/Jan/19
Find u_(min) and corresponding θ for this to be possible. (e is the coefficient of restitution for collision of ball with ground and with wall as well.)
$${Find}\:{u}_{{min}} \:{and}\:{corresponding}\:\theta\:{for}\: \\ $$$${this}\:{to}\:{be}\:{possible}. \\ $$$$\left({e}\:{is}\:{the}\:{coefficient}\:{of}\:{restitution}\right. \\ $$$${for}\:{collision}\:{of}\:{ball}\:{with}\:{ground}\: \\ $$$$\left.{and}\:{with}\:{wall}\:{as}\:{well}.\right) \\ $$
Answered by mr W last updated on 27/Jan/19
at point on ground: v_(1,x,i) =v_(1,x,f) =u cos θ v_(1,y,i) =u sin θ v_(1,y,f) =eu sin θ at point on wall: v_(2,x,i) =v_(1,x,f) =u cos θ t=(b/v_(2,x,f) )=(b/(u cos θ)) v_(2,y,i) =v_(1,y,f) −gt=eu sin θ−((gb)/(u cos θ)) h=v_(1,y,f) t−(1/2)gt^2 =eu sin θ×(b/(u cos θ))−(g/2)((b/(u cos θ)))^2 ⇒h=eb tan θ−((gb^2 )/(2u^2 cos^2 θ)) v_(2,x,f) =ev_(2,x,i) =eu cos θ v_(2,y,f) =v_(2,y,i) =eu sin θ−((gb)/(u cos θ)) at point on ground again: t=(b/v_(2,x,f) )=(b/(eu cos θ)) −h=v_(2,y,f) t−(1/2)gt^2 =(eu sin θ−((gb)/(u cos θ)))(b/(eu cos θ))−(g/2)((b/(eu cos θ)))^2 ⇒h=−b tan θ+((gb^2 )/(u^2 cos^2 θ))((1/e)+(1/(2e^2 ))) ⇒eb tan θ−((gb^2 )/(2u^2 cos^2 θ))=−b tan θ+((gb^2 )/(u^2 cos^2 θ))((1/e)+(1/(2e^2 ))) ⇒(e+1)tan θ=((gb)/(u^2 cos^2 θ))((1/2)+(1/e)+(1/(2e^2 ))) ⇒(e+1)tan θ=(((1+tan^2 θ)gb)/(2u^2 ))(((1+e)/e))^2 ⇒u^2 =((gb)/2)(((1+e)/e^2 ))(tan θ+(1/(tan θ))) since tan θ+(1/(tan θ))≥2 (“=”is at θ=45°) ⇒u_(min) ^2 =gb(((1+e)/e^2 )) ⇒u_(min) =(1/e)(√((1+e)gb)) at θ=45° ============ alternative way: after rebound from the ground and back: t=(b/(u cos θ))+(b/(eu cos θ))=(((1+e)b)/(eu cos θ)) 0=eu sin θ t−(1/2)gt^2 ⇒eu sin θ=(1/2)gt=(((1+e)gb)/(2eu cos θ)) ⇒u^2 =(((1+e)gb)/(e^2 sin 2θ)) ⇒u=(1/e)(√(((1+e)gb)/(sin 2θ))) ⇒u_(min) =((√((1+e)gb))/e) at θ=45°
$${at}\:{point}\:{on}\:{ground}: \\ $$$${v}_{\mathrm{1},{x},{i}} ={v}_{\mathrm{1},{x},{f}} ={u}\:\mathrm{cos}\:\theta \\ $$$${v}_{\mathrm{1},{y},{i}} ={u}\:\mathrm{sin}\:\theta \\ $$$${v}_{\mathrm{1},{y},{f}} ={eu}\:\mathrm{sin}\:\theta \\ $$$$ \\ $$$${at}\:{point}\:{on}\:{wall}: \\ $$$${v}_{\mathrm{2},{x},{i}} ={v}_{\mathrm{1},{x},{f}} ={u}\:\mathrm{cos}\:\theta \\ $$$${t}=\frac{{b}}{{v}_{\mathrm{2},{x},{f}} }=\frac{{b}}{{u}\:\mathrm{cos}\:\theta} \\ $$$${v}_{\mathrm{2},{y},{i}} ={v}_{\mathrm{1},{y},{f}} −{gt}={eu}\:\mathrm{sin}\:\theta−\frac{{gb}}{{u}\:\mathrm{cos}\:\theta} \\ $$$${h}={v}_{\mathrm{1},{y},{f}} {t}−\frac{\mathrm{1}}{\mathrm{2}}{gt}^{\mathrm{2}} ={eu}\:\mathrm{sin}\:\theta×\frac{{b}}{{u}\:\mathrm{cos}\:\theta}−\frac{{g}}{\mathrm{2}}\left(\frac{{b}}{{u}\:\mathrm{cos}\:\theta}\right)^{\mathrm{2}} \\ $$$$\Rightarrow{h}={eb}\:\mathrm{tan}\:\theta−\frac{{gb}^{\mathrm{2}} }{\mathrm{2}{u}^{\mathrm{2}} \mathrm{cos}^{\mathrm{2}} \:\theta} \\ $$$${v}_{\mathrm{2},{x},{f}} ={ev}_{\mathrm{2},{x},{i}} ={eu}\:\mathrm{cos}\:\theta \\ $$$${v}_{\mathrm{2},{y},{f}} ={v}_{\mathrm{2},{y},{i}} ={eu}\:\mathrm{sin}\:\theta−\frac{{gb}}{{u}\:\mathrm{cos}\:\theta} \\ $$$$ \\ $$$${at}\:{point}\:{on}\:{ground}\:{again}: \\ $$$${t}=\frac{{b}}{{v}_{\mathrm{2},{x},{f}} }=\frac{{b}}{{eu}\:\mathrm{cos}\:\theta} \\ $$$$−{h}={v}_{\mathrm{2},{y},{f}} {t}−\frac{\mathrm{1}}{\mathrm{2}}{gt}^{\mathrm{2}} =\left({eu}\:\mathrm{sin}\:\theta−\frac{{gb}}{{u}\:\mathrm{cos}\:\theta}\right)\frac{{b}}{{eu}\:\mathrm{cos}\:\theta}−\frac{{g}}{\mathrm{2}}\left(\frac{{b}}{{eu}\:\mathrm{cos}\:\theta}\right)^{\mathrm{2}} \\ $$$$\Rightarrow{h}=−{b}\:\mathrm{tan}\:\theta+\frac{{gb}^{\mathrm{2}} }{{u}^{\mathrm{2}} \:\mathrm{cos}^{\mathrm{2}} \:\theta}\left(\frac{\mathrm{1}}{{e}}+\frac{\mathrm{1}}{\mathrm{2}{e}^{\mathrm{2}} }\right) \\ $$$$ \\ $$$$\Rightarrow{eb}\:\mathrm{tan}\:\theta−\frac{{gb}^{\mathrm{2}} }{\mathrm{2}{u}^{\mathrm{2}} \mathrm{cos}^{\mathrm{2}} \:\theta}=−{b}\:\mathrm{tan}\:\theta+\frac{{gb}^{\mathrm{2}} }{{u}^{\mathrm{2}} \:\mathrm{cos}^{\mathrm{2}} \:\theta}\left(\frac{\mathrm{1}}{{e}}+\frac{\mathrm{1}}{\mathrm{2}{e}^{\mathrm{2}} }\right) \\ $$$$\Rightarrow\left({e}+\mathrm{1}\right)\mathrm{tan}\:\theta=\frac{{gb}}{{u}^{\mathrm{2}} \:\mathrm{cos}^{\mathrm{2}} \:\theta}\left(\frac{\mathrm{1}}{\mathrm{2}}+\frac{\mathrm{1}}{{e}}+\frac{\mathrm{1}}{\mathrm{2}{e}^{\mathrm{2}} }\right) \\ $$$$\Rightarrow\left({e}+\mathrm{1}\right)\mathrm{tan}\:\theta=\frac{\left(\mathrm{1}+\mathrm{tan}^{\mathrm{2}} \:\theta\right){gb}}{\mathrm{2}{u}^{\mathrm{2}} }\left(\frac{\mathrm{1}+{e}}{{e}}\right)^{\mathrm{2}} \\ $$$$\Rightarrow{u}^{\mathrm{2}} =\frac{{gb}}{\mathrm{2}}\left(\frac{\mathrm{1}+{e}}{{e}^{\mathrm{2}} }\right)\left(\mathrm{tan}\:\theta+\frac{\mathrm{1}}{\mathrm{tan}\:\theta}\right) \\ $$$${since}\:\mathrm{tan}\:\theta+\frac{\mathrm{1}}{\mathrm{tan}\:\theta}\geqslant\mathrm{2}\:\left(“=''{is}\:{at}\:\theta=\mathrm{45}°\right) \\ $$$$\Rightarrow{u}_{{min}} ^{\mathrm{2}} ={gb}\left(\frac{\mathrm{1}+{e}}{{e}^{\mathrm{2}} }\right) \\ $$$$\Rightarrow{u}_{{min}} =\frac{\mathrm{1}}{{e}}\sqrt{\left(\mathrm{1}+{e}\right){gb}} \\ $$$${at}\:\theta=\mathrm{45}° \\ $$$$ \\ $$$$============ \\ $$$${alternative}\:{way}: \\ $$$${after}\:{rebound}\:{from}\:{the}\:{ground}\:{and} \\ $$$${back}: \\ $$$${t}=\frac{{b}}{{u}\:\mathrm{cos}\:\theta}+\frac{{b}}{{eu}\:\mathrm{cos}\:\theta}=\frac{\left(\mathrm{1}+{e}\right){b}}{{eu}\:\mathrm{cos}\:\theta} \\ $$$$\mathrm{0}={eu}\:\mathrm{sin}\:\theta\:{t}−\frac{\mathrm{1}}{\mathrm{2}}{gt}^{\mathrm{2}} \\ $$$$\Rightarrow{eu}\:\mathrm{sin}\:\theta=\frac{\mathrm{1}}{\mathrm{2}}{gt}=\frac{\left(\mathrm{1}+{e}\right){gb}}{\mathrm{2}{eu}\:\mathrm{cos}\:\theta} \\ $$$$\Rightarrow{u}^{\mathrm{2}} =\frac{\left(\mathrm{1}+{e}\right){gb}}{{e}^{\mathrm{2}} \:\mathrm{sin}\:\mathrm{2}\theta} \\ $$$$\Rightarrow{u}=\frac{\mathrm{1}}{{e}}\sqrt{\frac{\left(\mathrm{1}+{e}\right){gb}}{\mathrm{sin}\:\mathrm{2}\theta}} \\ $$$$\Rightarrow{u}_{{min}} =\frac{\sqrt{\left(\mathrm{1}+{e}\right){gb}}}{{e}}\:{at}\:\theta=\mathrm{45}° \\ $$
Commented by ajfour last updated on 26/Jan/19
thanks Sir. Beautiful result, isn′t ?
$${thanks}\:\:{Sir}.\:\mathcal{B}{eautiful}\:{result},\:{isn}'{t}\:? \\ $$
Commented by mr W last updated on 26/Jan/19
it′s indeed a very nice question!
$${it}'{s}\:{indeed}\:{a}\:{very}\:{nice}\:{question}! \\ $$
Commented by mr W last updated on 26/Jan/19
that′s it! thank you for checking sir!
$${that}'{s}\:{it}!\:{thank}\:{you}\:{for}\:{checking}\:{sir}! \\ $$
Commented by mr W last updated on 26/Jan/19
as children we often tried to play this with a tennis ball to see if we could get the ball back to the same point on the ground.
$${as}\:{children}\:{we}\:{often}\:{tried}\:{to}\:{play} \\ $$$${this}\:{with}\:{a}\:{tennis}\:{ball}\:{to}\:{see}\:{if}\:{we}\:{could} \\ $$$${get}\:{the}\:{ball}\:{back}\:{to}\:{the}\:{same}\:{point}\:{on}\:{the} \\ $$$${ground}. \\ $$

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