Question Number 223653 by ajfour last updated on 01/Aug/25
Commented by ajfour last updated on 01/Aug/25
$${Find}\:\theta\:{and}\:{k}. \\ $$$${We}\:{need}\:{to}\:{just}\:{make}\:{the}\:{ball}\:{make} \\ $$$${it}\:{to}\:{the}\:{cliff}.\:{The}\:{ground}\:{is}\:{frictionless}. \\ $$
Answered by mr W last updated on 02/Aug/25
Commented by mr W last updated on 08/Aug/25
Commented by mr W last updated on 08/Aug/25
Commented by ajfour last updated on 07/Aug/25
$${Thank}\:{you}\:{sir}.\:{Any}\:{answer}\:{for}\:\theta\:{as}\:{an}\:{example}.. \\ $$
Commented by mr W last updated on 08/Aug/25
![tan α=((2(H−R sin θ))/(a+R(1−cos θ))) mv cos α=MV ⇒V=((mv cos α)/M)=μv cos α with μ=(m/M) v cos (θ−α)+V cos θ=ev_1 sin θ ...(i) v sin (θ−α)+V sin θ=v_1 cos θ ...(ii) ((v cos (θ−α)+V cos θ)/(v sin (θ−α)+V sin θ))=e tan θ ((cos θ cos α+sin θ sin α+μ cos α cos θ)/(sin θ cos α−cos θ sin α+μ cos α sin θ))=e tan θ ((1+μ+tan θ tan α)/((1+μ) tan θ−tan α))=e tan θ tan α=(((1+μ)(e tan^2 θ−1))/((1+e) tan θ)) ⇒(((1+μ)(e tan^2 θ−1))/((1+e) tan θ))=((2(H−R sin θ))/(a+R(1−cos θ))) v sin α=(√(2g(H−R sin θ))) ⇒v=((√(2g(H−R sin θ)))/(sin α)) ⇒u=v cos α=((a+R(1−cos θ))/( (√((2(H−R sin θ))/g)))) v_1 =[((sin (θ−α))/(cos θ))+μ cos α tan θ]v ⇒v_1 =[(((1+μ)tan θ)/(tan α))−1](√(2g(H−R sin θ))) 2g(k−R sin θ)=v_1 ^2 ⇒k=R sin θ+[(((1+μ)tan θ)/(tan α))−1]^2 (H−R sin θ)](https://www.tinkutara.com/question/Q223870.png)
$$\mathrm{tan}\:\alpha=\frac{\mathrm{2}\left({H}−{R}\:\mathrm{sin}\:\theta\right)}{{a}+{R}\left(\mathrm{1}−\mathrm{cos}\:\theta\right)} \\ $$$${mv}\:\mathrm{cos}\:\alpha={MV}\: \\ $$$$\Rightarrow{V}=\frac{{mv}\:\mathrm{cos}\:\alpha}{{M}}=\mu{v}\:\mathrm{cos}\:\alpha \\ $$$${with}\:\mu=\frac{{m}}{{M}} \\ $$$${v}\:\mathrm{cos}\:\left(\theta−\alpha\right)+{V}\:\mathrm{cos}\:\theta={ev}_{\mathrm{1}} \:\mathrm{sin}\:\theta\:\:\:…\left({i}\right) \\ $$$${v}\:\mathrm{sin}\:\left(\theta−\alpha\right)+{V}\:\mathrm{sin}\:\theta={v}_{\mathrm{1}} \:\mathrm{cos}\:\theta\:\:\:…\left({ii}\right) \\ $$$$\frac{{v}\:\mathrm{cos}\:\left(\theta−\alpha\right)+{V}\:\mathrm{cos}\:\theta}{{v}\:\mathrm{sin}\:\left(\theta−\alpha\right)+{V}\:\mathrm{sin}\:\theta}={e}\:\mathrm{tan}\:\theta \\ $$$$\frac{\mathrm{cos}\:\theta\:\mathrm{cos}\:\alpha+\mathrm{sin}\:\theta\:\mathrm{sin}\:\alpha+\mu\:\mathrm{cos}\:\alpha\:\mathrm{cos}\:\theta}{\mathrm{sin}\:\theta\:\mathrm{cos}\:\alpha−\mathrm{cos}\:\theta\:\mathrm{sin}\:\alpha+\mu\:\mathrm{cos}\:\alpha\:\mathrm{sin}\:\theta}={e}\:\mathrm{tan}\:\theta \\ $$$$\frac{\mathrm{1}+\mu+\mathrm{tan}\:\theta\:\mathrm{tan}\:\alpha}{\left(\mathrm{1}+\mu\right)\:\mathrm{tan}\:\theta−\mathrm{tan}\:\alpha}={e}\:\mathrm{tan}\:\theta \\ $$$$\mathrm{tan}\:\alpha=\frac{\left(\mathrm{1}+\mu\right)\left({e}\:\mathrm{tan}^{\mathrm{2}} \:\theta−\mathrm{1}\right)}{\left(\mathrm{1}+{e}\right)\:\mathrm{tan}\:\theta} \\ $$$$\Rightarrow\frac{\left(\mathrm{1}+\mu\right)\left({e}\:\mathrm{tan}^{\mathrm{2}} \:\theta−\mathrm{1}\right)}{\left(\mathrm{1}+{e}\right)\:\mathrm{tan}\:\theta}=\frac{\mathrm{2}\left({H}−{R}\:\mathrm{sin}\:\theta\right)}{{a}+{R}\left(\mathrm{1}−\mathrm{cos}\:\theta\right)} \\ $$$${v}\:\mathrm{sin}\:\alpha=\sqrt{\mathrm{2}{g}\left({H}−{R}\:\mathrm{sin}\:\theta\right)} \\ $$$$\Rightarrow{v}=\frac{\sqrt{\mathrm{2}{g}\left({H}−{R}\:\mathrm{sin}\:\theta\right)}}{\mathrm{sin}\:\alpha} \\ $$$$\Rightarrow{u}={v}\:\mathrm{cos}\:\alpha=\frac{{a}+{R}\left(\mathrm{1}−\mathrm{cos}\:\theta\right)}{\:\sqrt{\frac{\mathrm{2}\left({H}−{R}\:\mathrm{sin}\:\theta\right)}{{g}}}} \\ $$$${v}_{\mathrm{1}} =\left[\frac{\mathrm{sin}\:\left(\theta−\alpha\right)}{\mathrm{cos}\:\theta}+\mu\:\mathrm{cos}\:\alpha\:\mathrm{tan}\:\theta\right]{v} \\ $$$$\Rightarrow{v}_{\mathrm{1}} =\left[\frac{\left(\mathrm{1}+\mu\right)\mathrm{tan}\:\theta}{\mathrm{tan}\:\alpha}−\mathrm{1}\right]\sqrt{\mathrm{2}{g}\left({H}−{R}\:\mathrm{sin}\:\theta\right)} \\ $$$$\mathrm{2}{g}\left({k}−{R}\:\mathrm{sin}\:\theta\right)={v}_{\mathrm{1}} ^{\mathrm{2}} \\ $$$$\Rightarrow{k}={R}\:\mathrm{sin}\:\theta+\left[\frac{\left(\mathrm{1}+\mu\right)\mathrm{tan}\:\theta}{\mathrm{tan}\:\alpha}−\mathrm{1}\right]^{\mathrm{2}} \left({H}−{R}\:\mathrm{sin}\:\theta\right) \\ $$
Commented by ajfour last updated on 08/Aug/25
$${yes},\:{i}\:{see}\:{it}.\:{Thank}\:{you}\:{sir}. \\ $$
Commented by mr W last updated on 08/Aug/25
$$\left({i}\right)\:{and}\:\left({ii}\right)\:{are}\:{right},\:{sir}? \\ $$