Question Number 209529 by mr W last updated on 13/Jul/24
![](https://www.tinkutara.com/question/34420.png)
Commented by mr W last updated on 13/Jul/24
![find u_(min) such that the ball can fly out of the well. (collisions with the wall are elastic)](https://www.tinkutara.com/question/Q209530.png)
$${find}\:{u}_{{min}} \:{such}\:{that}\:{the}\:{ball}\:{can}\:{fly} \\ $$$${out}\:{of}\:{the}\:{well}. \\ $$$$\left({collisions}\:{with}\:{the}\:{wall}\:{are}\:{elastic}\right)\: \\ $$
Answered by mr W last updated on 13/Jul/24
![](https://www.tinkutara.com/question/34422.png)
Commented by mr W last updated on 13/Jul/24
![nb=u cos θ t ⇒t=((nb)/(u cos θ)) u sin θt−((gt^2 )/2)≥h nb tan θ−((gn^2 b^2 (1+tan^2 θ))/(2u^2 ))≥h ((2u^2 )/(gh))≥((1+tan^2 θ)/((tan θ−(h/(nb)))(h/(nb)))) let t=tan θ, λ=(h/(nb)) Φ=((1+t^2 )/((t−λ)λ)) (dΦ/dt)=((2t)/((t−λ)λ))−(((1+t^2 ))/((t−λ)^2 λ))=0 ⇒t^2 −2λt−1=0 ⇒t=λ+(√(λ^2 +1)) Φ_(min) =((2(λ^2 +λ(√(λ^2 +1))+1))/(λ(√(λ^2 +1))))=2(1+(√(1+(1/λ)))) ⇒u_(min) =(√(gh(1+(√(1+((nb)/h))))))](https://www.tinkutara.com/question/Q209535.png)
$${nb}={u}\:\mathrm{cos}\:\theta\:{t}\:\Rightarrow{t}=\frac{{nb}}{{u}\:\mathrm{cos}\:\theta} \\ $$$${u}\:\mathrm{sin}\:\theta{t}−\frac{{gt}^{\mathrm{2}} }{\mathrm{2}}\geqslant{h} \\ $$$${nb}\:\mathrm{tan}\:\theta−\frac{{gn}^{\mathrm{2}} {b}^{\mathrm{2}} \left(\mathrm{1}+\mathrm{tan}^{\mathrm{2}} \:\theta\right)}{\mathrm{2}{u}^{\mathrm{2}} }\geqslant{h} \\ $$$$\frac{\mathrm{2}{u}^{\mathrm{2}} }{{gh}}\geqslant\frac{\mathrm{1}+\mathrm{tan}^{\mathrm{2}} \:\theta}{\left(\mathrm{tan}\:\theta−\frac{{h}}{{nb}}\right)\frac{{h}}{{nb}}} \\ $$$${let}\:{t}=\mathrm{tan}\:\theta,\:\lambda=\frac{{h}}{{nb}} \\ $$$$\Phi=\frac{\mathrm{1}+{t}^{\mathrm{2}} }{\left({t}−\lambda\right)\lambda} \\ $$$$\frac{{d}\Phi}{{dt}}=\frac{\mathrm{2}{t}}{\left({t}−\lambda\right)\lambda}−\frac{\left(\mathrm{1}+{t}^{\mathrm{2}} \right)}{\left({t}−\lambda\right)^{\mathrm{2}} \lambda}=\mathrm{0} \\ $$$$\Rightarrow{t}^{\mathrm{2}} −\mathrm{2}\lambda{t}−\mathrm{1}=\mathrm{0} \\ $$$$\Rightarrow{t}=\lambda+\sqrt{\lambda^{\mathrm{2}} +\mathrm{1}} \\ $$$$\Phi_{{min}} =\frac{\mathrm{2}\left(\lambda^{\mathrm{2}} +\lambda\sqrt{\lambda^{\mathrm{2}} +\mathrm{1}}+\mathrm{1}\right)}{\lambda\sqrt{\lambda^{\mathrm{2}} +\mathrm{1}}}=\mathrm{2}\left(\mathrm{1}+\sqrt{\mathrm{1}+\frac{\mathrm{1}}{\lambda}}\right) \\ $$$$\Rightarrow{u}_{{min}} =\sqrt{{gh}\left(\mathrm{1}+\sqrt{\mathrm{1}+\frac{{nb}}{{h}}}\right)} \\ $$