Question-132102

Question Number 132102 by mr W last updated on 14/Feb/21

Commented bymr W last updated on 16/Feb/21

$a b a l l i s t h r o w n f r o m p o i n t A w i t h$
$s p e e d u a n d s t r i k e s a t a p o i n t B$
$o n t h e s e m i s p h e r i c a l s u r f a c e w i t h$
$r a d i u s R a n d r e t u r n s b a c k t o p o i n t A .$
$i f t h e r e s t i t u t i o n c o e f f i c i e n t o f$
$t h e c o l l i s i o n i s e, f i n d t h e m i n i m u m$
$s p e e d u t h e b a l l m u s t h a v e a n d a t$
$w h i c h a n g l e θ t h e b a l l s h o u l d b e$
$t h r o w n .$

Commented byotchereabdullai@gmail.com last updated on 14/Feb/21

$You are the world best no one can$
$challenge you!$

Commented byajfour last updated on 15/Feb/21

$For u_(min) I got φ=30° t_1 =(√((2(√3)R)/g)){1±(√((1/e^2 )−1))} u_(min) ^2 =(((R(√3))/(2t))+((gt)/4))^2 +((3(√3))/2)gR((1/e^2 )−1)$

$F o r u_{m i n} I g o t ϕ = 30 °$
$t_{1} = \sqrt{\frac{2 \sqrt{3} R}{g}} {1 \pm \sqrt{\frac{1}{e^{2}} - 1}}$
$u_{m i n}^{2} = {(\frac{R \sqrt{3}}{2 t} + \frac{g t}{4})}^{2} + \frac{3 \sqrt{3}}{2} g R (\frac{1}{e^{2}} - 1)$

Commented bymr W last updated on 15/Feb/21

$d o y o u m e a n t h a t u_{m i n} i s a l w a y s a t$
$ϕ = 30 ° i n d e p e n t e n t l y f r o m e ?$
$i {c a n}^{'} t c o n f i r m t h i s s i r .$

Answered by mr W last updated on 25/Feb/21

Commented bymr W last updated on 12/Feb/21

Commented bymr W last updated on 16/Feb/21

$motion from A to B: t=((R(1+cos φ))/(u cos θ)) R sin φ=−u sin θ×((R(1+cos φ))/(u cos θ))+(g/2)×((R^2 (1+cos φ)^2 )/(u^2 cos^2 θ)) ((sin φ)/(1+cos φ))=((gR)/(2u^2 ))×(1+cos φ)(1+tan^2 θ)−tan θ let λ=((gR)/(2u^2 )) ((sin φ)/(1+cos φ))=λ(1+cos φ)(1+tan^2 θ)−tan θ λ(1+cos φ)tan^2 θ−tan θ+((λ(1+cos φ)^2 −sin φ)/(1+cos φ))=0 tan θ=((1±(√(1+4λ sin φ−4λ^2 (1+cos φ)^2 )))/(2λ(1+cos φ))) ⇒θ=tan^(−1) [((1±(√(1+4λ sin φ−4λ^2 (1+cos φ)^2 )))/(2λ(1+cos φ)))] at point B: U_x =u cos θ U_y =gt−u sin θ=((gR(1+cos φ))/(u cos θ))−u sin θ U_⊥ =U_x cos φ+U_y sin φ U_⊥ =u cos θcos φ+[((gR(1+cos φ))/(u cos θ))−u sin θ]sin φ ⇒U_⊥ =u cos (φ+θ)+((gR(1+cos φ)sin φ)/(u cos θ)) U_∥ =−U_x sin φ+U_y cos φ U_∥ =−u cos θ sin φ+[((gR(1+cos φ))/(u cos θ))−u sin θ]cos φ ⇒U_∥ =−u sin (φ+θ)+((gR(1+cos φ)cos φ)/(u cos θ)) V_⊥ =eU_⊥ ⇒V_⊥ =eu cos (φ+θ)+((egR(1+cos φ)sin φ)/(u cos θ)) V_∥ =U_∥ ⇒V_∥ =−u sin (φ+θ)+((gR(1+cos φ)cos φ)/(u cos θ)) V_x =V_⊥ cos φ+V_∥ sin φ V_x =[eu cos (φ+θ)+((egR(1+cos φ)sin φ)/(u cos θ))]cos φ+[−u sin (φ+θ)+((gR(1+cos φ)cos φ)/(u cos θ))]sin φ V_x =eu cos (φ+θ)cos φ+((egR(1+cos φ)sin φ cos φ)/(u cos θ))−u sin (φ+θ)sin φ+((gR(1+cos φ)sin φcos φ)/(u cos θ)) ⇒V_x =u[(1+e)cos (φ+θ)cos φ−cos θ+((λ(1+e)(1+cos φ) sin 2φ)/(cos θ))] V_y =V_⊥ sin φ−V_∥ cos φ V_y =[eu cos (φ+θ)+((egR(1+cos φ)sin φ)/(u cos θ))]sin φ−[−u sin (φ+θ)+((gR(1+cos φ)cos φ)/(u cos θ))]cos φ ⇒V_y =u[(1+e)cos (φ+θ)sin φ+sin θ+((2λ(1+cos φ)((1+e) sin^2 φ−1))/(cos θ))] return from B to A: t=((R(1+cos φ))/((1+e)cos (φ+θ)cos φ−cos θ+((λ(1+cos φ)(1+e) sin 2φ)/(cos θ))))×(1/u) ⇒((tu)/R)=ξ=((1+cos φ)/((1+e)cos (φ+θ)cos φ−cos θ+((λ(1+e)(1+cos φ) sin 2φ)/(cos θ)))) R sin φ=t{u[(1+e)cos (φ+θ)sin φ+sin θ]+((2λ(1+cos φ)(e sin^2 φ−cos^2 φ))/(cos θ))−((gt)/2)} sin φ=((tu)/R)[(1+e)cos (φ+θ)sin φ+sin θ+((2λ(1+cos φ)(e sin^2 φ−cos^2 φ))/(cos θ))−((λtu)/R)] sin φ=ξ[(1+e)cos (φ+θ)sin φ+sin θ+((2λ(1+cos φ)((1+e)sin^2 φ−1))/(cos θ))−λξ] or sin φ=ξ[e′cos (φ+θ) sin φ+sin θ+((2λ(1+cos φ)(e′sin^2 φ−1))/(cos θ))−λξ] ...(I) with e′=1+e λ=((gR)/(2u^2 )) θ=tan^(−1) [((1+(√(1+4λ sin φ−4λ^2 (1+cos φ)^2 )))/(2λ(1+cos φ)))] ∗) ξ=((1+cos φ)/(e′cos (φ+θ) cos φ−cos θ+((λe′(1+cos φ) sin 2φ)/(cos θ)))) for a given e ∈(0,1] we can find solution(s) for φ in (I) for some values of parameter λ. the maximum value of λ such that a solution for φ exists can be determined numerically. this corresponds to the minimum speed the ball must have: u_(min) =(√((gR)/(2λ_(max) ))) examples: e=0.5 ⇒u_(min) ≈2.7682(√(gR)) e=0.75 ⇒u_(min) ≈1.6082(√(gR)) e=1 ⇒u_(min) ≈0.9098(√(gR)) (= Q65589) ∗) with the other solution θ=tan^(−1) [((1−(√(1+4λ sin φ−4λ^2 (1+cos φ)^2 )))/(2λ(1+cos φ)))] we can get a solution for the case that the ball strikes the surface upwards, i.e. counterclockwise. but in this case more energy is needed, i.e. u_(min) is larger than with θ=tan^(−1) [((1+(√(1+4λ sin φ−4λ^2 (1+cos φ)^2 )))/(2λ(1+cos φ)))] see examples with e=0.5 and 0.75.$

$\underset{―}{m o t i o n f r o m A t o B :}$
$t = \frac{R (1 + \cos ϕ)}{u \cos θ}$
$R \sin ϕ = - u \sin θ \times \frac{R (1 + \cos ϕ)}{u \cos θ} + \frac{g}{2} \times \frac{R^{2} {(1 + \cos ϕ)}^{2}}{u^{2} \cos^{2} θ}$
$\frac{\sin ϕ}{1 + \cos ϕ} = \frac{g R}{2 u^{2}} \times (1 + \cos ϕ) (1 + \tan^{2} θ) - \tan θ$
$l e t λ = \frac{g R}{2 u^{2}}$
$\frac{\sin ϕ}{1 + \cos ϕ} = λ (1 + \cos ϕ) (1 + \tan^{2} θ) - \tan θ$
$λ (1 + \cos ϕ) \tan^{2} θ - \tan θ + \frac{λ {(1 + \cos ϕ)}^{2} - \sin ϕ}{1 + \cos ϕ} = 0$
$\tan θ = \frac{1 \pm \sqrt{1 + 4 λ \sin ϕ - 4 λ^{2} {(1 + \cos ϕ)}^{2}}}{2 λ (1 + \cos ϕ)}$
$\Rightarrow θ = \tan^{- 1} [\frac{1 \pm \sqrt{1 + 4 λ \sin ϕ - 4 λ^{2} {(1 + \cos ϕ)}^{2}}}{2 λ (1 + \cos ϕ)}]$

$\underset{―}{a t p o i n t B :}$
$U_{x} = u \cos θ$
$U_{y} = g t - u \sin θ = \frac{g R (1 + \cos ϕ)}{u \cos θ} - u \sin θ$
$U_{⊥} = U_{x} \cos ϕ + U_{y} \sin ϕ$
$U_{⊥} = u \cos θ \cos ϕ + [\frac{g R (1 + \cos ϕ)}{u \cos θ} - u \sin θ] \sin ϕ$
$\Rightarrow U_{⊥} = u \cos (ϕ + θ) + \frac{g R (1 + \cos ϕ) \sin ϕ}{u \cos θ}$
$U_{∥} = - U_{x} \sin ϕ + U_{y} \cos ϕ$
$U_{∥} = - u \cos θ \sin ϕ + [\frac{g R (1 + \cos ϕ)}{u \cos θ} - u \sin θ] \cos ϕ$
$\Rightarrow U_{∥} = - u \sin (ϕ + θ) + \frac{g R (1 + \cos ϕ) \cos ϕ}{u \cos θ}$

$V_{⊥} = {e U}_{⊥}$
$\Rightarrow V_{⊥} = e u \cos (ϕ + θ) + \frac{e g R (1 + \cos ϕ) \sin ϕ}{u \cos θ}$
$V_{∥} = U_{∥}$
$\Rightarrow V_{∥} = - u \sin (ϕ + θ) + \frac{g R (1 + \cos ϕ) \cos ϕ}{u \cos θ}$

$V_{x} = V_{⊥} \cos ϕ + V_{∥} \sin ϕ$
$V_{x} = [e u \cos (ϕ + θ) + \frac{e g R (1 + \cos ϕ) \sin ϕ}{u \cos θ}] \cos ϕ + [- u \sin (ϕ + θ) + \frac{g R (1 + \cos ϕ) \cos ϕ}{u \cos θ}] \sin ϕ$
$V_{x} = e u \cos (ϕ + θ) \cos ϕ + \frac{e g R (1 + \cos ϕ) \sin ϕ \cos ϕ}{u \cos θ} - u \sin (ϕ + θ) \sin ϕ + \frac{g R (1 + \cos ϕ) \sin ϕ \cos ϕ}{u \cos θ}$
$\Rightarrow V_{x} = u [(1 + e) \cos (ϕ + θ) \cos ϕ - \cos θ + \frac{λ (1 + e) (1 + \cos ϕ) \sin 2 ϕ}{\cos θ}]$
$V_{y} = V_{⊥} \sin ϕ - V_{∥} \cos ϕ$
$V_{y} = [e u \cos (ϕ + θ) + \frac{e g R (1 + \cos ϕ) \sin ϕ}{u \cos θ}] \sin ϕ - [- u \sin (ϕ + θ) + \frac{g R (1 + \cos ϕ) \cos ϕ}{u \cos θ}] \cos ϕ$
$\Rightarrow V_{y} = u [(1 + e) \cos (ϕ + θ) \sin ϕ + \sin θ + \frac{2 λ (1 + \cos ϕ) ((1 + e) \sin^{2} ϕ - 1)}{\cos θ}]$

$\underset{―}{r e t u r n f r o m B t o A :}$
$t = \frac{R (1 + \cos ϕ)}{(1 + e) \cos (ϕ + θ) \cos ϕ - \cos θ + \frac{λ (1 + \cos ϕ) (1 + e) \sin 2 ϕ}{\cos θ}} \times \frac{1}{u}$
$\Rightarrow \frac{t u}{R} = ξ = \frac{1 + \cos ϕ}{(1 + e) \cos (ϕ + θ) \cos ϕ - \cos θ + \frac{λ (1 + e) (1 + \cos ϕ) \sin 2 ϕ}{\cos θ}}$
$R \sin ϕ = t {u [(1 + e) \cos (ϕ + θ) \sin ϕ + \sin θ] + \frac{2 λ (1 + \cos ϕ) (e \sin^{2} ϕ - \cos^{2} ϕ)}{\cos θ} - \frac{g t}{2}}$
$\sin ϕ = \frac{t u}{R} [(1 + e) \cos (ϕ + θ) \sin ϕ + \sin θ + \frac{2 λ (1 + \cos ϕ) (e \sin^{2} ϕ - \cos^{2} ϕ)}{\cos θ} - \frac{λ t u}{R}]$
$\sin ϕ = ξ [(1 + e) \cos (ϕ + θ) \sin ϕ + \sin θ + \frac{2 λ (1 + \cos ϕ) ((1 + e) \sin^{2} ϕ - 1)}{\cos θ} - λ ξ]$
$o r$
$\sin ϕ = ξ [e^{'} \cos (ϕ + θ) \sin ϕ + \sin θ + \frac{2 λ (1 + \cos ϕ) (e^{'} \sin^{2} ϕ - 1)}{\cos θ} - λ ξ] \dots (I)$
$w i t h$
$e^{'} = 1 + e$
$λ = \frac{g R}{2 u^{2}}$
$θ = \tan^{- 1} [\frac{1 + \sqrt{1 + 4 λ \sin ϕ - 4 λ^{2} {(1 + \cos ϕ)}^{2}}}{2 λ (1 + \cos ϕ)}] *)$
$ξ = \frac{1 + \cos ϕ}{e^{'} \cos (ϕ + θ) \cos ϕ - \cos θ + \frac{λ e^{'} (1 + \cos ϕ) \sin 2 ϕ}{\cos θ}}$
$f o r a g i v e n e \in (0, 1] w e c a n f i n d$
$s o l u t i o n (s) f o r ϕ i n (I) f o r s o m e$
$v a l u e s o f p a r a m e t e r λ . t h e m a x i m u m$
$v a l u e o f λ s u c h t h a t a s o l u t i o n f o r ϕ$
$e x i s t s c a n b e d e t e r m i n e d n u m e r i c a l l y .$
$t h i s c o r r e s p o n d s t o t h e m i n i m u m$
$s p e e d t h e b a l l m u s t h a v e :$
$u_{m i n} = \sqrt{\frac{g R}{2 λ_{m a x}}}$

$e x a m p l e s :$
$e = 0.5 \Rightarrow u_{m i n} \approx 2.7682 \sqrt{g R}$
$e = 0.75 \Rightarrow u_{m i n} \approx 1.6082 \sqrt{g R}$
$e = 1 \Rightarrow u_{m i n} \approx 0.9098 \sqrt{g R} (= Q 65589)$

$*)$
$w i t h t h e o t h e r s o l u t i o n$
$θ = \tan^{- 1} [\frac{1 - \sqrt{1 + 4 λ \sin ϕ - 4 λ^{2} {(1 + \cos ϕ)}^{2}}}{2 λ (1 + \cos ϕ)}]$
$w e c a n g e t a s o l u t i o n f o r t h e c a s e$
$t h a t t h e b a l l s t r i k e s t h e s u r f a c e$
$u p w a r d s, i . e . c o u n t e r c l o c k w i s e .$
$b u t i n t h i s c a s e m o r e e n e r g y i s$
$n e e d e d, i . e . u_{m i n} i s l a r g e r t h a n w i t h$
$θ = \tan^{- 1} [\frac{1 + \sqrt{1 + 4 λ \sin ϕ - 4 λ^{2} {(1 + \cos ϕ)}^{2}}}{2 λ (1 + \cos ϕ)}]$
$s e e e x a m p l e s w i t h e = 0.5 a n d 0.75 .$

Commented bymr W last updated on 15/Feb/21

Commented bymr W last updated on 13/Feb/21

Commented bymr W last updated on 13/Feb/21

Commented bymr W last updated on 12/Feb/21

Commented bymr W last updated on 12/Feb/21

Commented bymr W last updated on 14/Feb/21

$f o l l o w i n g d i a g r a m s s h o w d i f f e r e n t$
$s i t u a t i o n s f o r e = 0.75 .$

$g e n e r a l l y :$
$f o r a g i v e n e \in (0, 1] t h e r e e x i s t s a n$
$u_{m i n} . f o r a n y u > u_{m i n} t h e r e a r e f o u r$
$p o s s i b l e w a y s h o w t h e b a l l r e t u r n s$
$b a c k t o i t s s t a r t i n g p o s i t i o n : t w o$
$w a y s c l o c k w i s e a n d t w o c o u n t e r -$
$c l o c k w i s e .$

Commented bymr W last updated on 15/Feb/21

Commented bymr W last updated on 13/Feb/21

Commented bymr W last updated on 14/Feb/21

Commented bymr W last updated on 13/Feb/21

Commented bymr W last updated on 14/Feb/21

Answered by mr W last updated on 14/Feb/21

Commented bymr W last updated on 14/Feb/21

$b a c k g r o u n d k n o w l e d g e$

$w h a t h a p p e n s w h e n a b a l l s t r i k e s o n$
$a w a l l a n d t h e r e s t i t u t i o n c o e f f i c i e n t$
$o f t h e c o l l i s i o n i s e ?$

$U = c o m i n g s p e e d$
$V = l e a v i n g s p e e d$

$U_{⊥} = U \cos δ_{1}$
$U_{∥} = U \sin δ_{1}$
$V_{⊥} = V \cos δ_{2}$
$V_{∥} = V \sin δ_{2}$

$V_{⊥} = e \times U_{⊥} \Rightarrow V \cos δ_{2} = e \times U \cos δ_{1}$
$V_{∥} = U_{∥} \Rightarrow V \sin δ_{2} = U \sin δ_{1}$
$\tan δ_{2} = \frac{\tan δ_{1}}{e}$
$δ_{2} ⩾ δ_{1}$

${K E}_{1} = \frac{1}{2} {m U}^{2}$
${K E}_{2} = \frac{1}{2} {m V}^{2} = \frac{1}{2} {m U}^{2} (\sin^{2} δ_{1} + e^{2} \cos^{2} δ_{1})$
$\Rightarrow {K E}_{2} = (\sin^{2} δ_{1} + e^{2} \cos^{2} δ_{1}) {K E}_{1}$
$w e s e e {K E}_{2} = {K E}_{1} o n l y w h e n e = 1,$
$o t h e r w i s e {K E}_{2} < {K E}_{1} .$

$t h e c o l l i s i o n i s c a l l e d e l a s t i c i f e = 1 .$

Commented byAr Brandon last updated on 13/Feb/21

Wow ! Where did you learn your geometry Sir ? I feel you have this unique skill here. \n

Commented byAr Brandon last updated on 14/Feb/21

\n

Commented bymr W last updated on 14/Feb/21

$t h e b e s t t e a c h e r i s t h e o w n i n t e r e s t \dots$

Answered by ajfour last updated on 15/Feb/21

Commented byajfour last updated on 15/Feb/21

$F i r s t j u s t a s s u m i n g (w i t h o u t$
$p r o o f) t h a t u i s u_{m i n} w h e n$
$l e a s t e n e r g y i s l o s t u p o n c o l l i s i o n .$
$L e t j u s t b e f o r e c o l l i s i o n$
$v e l o c i t y {\bar{v}}_{1} = - q \hat{i} - p \hat{j}$
$a n d j u s t a f t e r {\bar{v}}_{2} = - q \hat{i} + e p \hat{j}$
$- △ K = \frac{p^{2}}{2} (1 - e^{2})$
$T h i s i s m i n i m u m i f p i s$
$m i n i m u m .$
$B u t b a l l h a s t o r e a c h b a c k \Rightarrow$
${(e^{2} p^{2})}_{m i n} = 2 (g \sin 2 ϕ) R (1 + \cos 2 ϕ)$
$s a y 2 ϕ = θ, a n d$
$f (θ) = \sin θ (1 + \cos θ)$
$f^{'} (θ) = \cos θ + \cos^{2} θ - \sin^{2} θ$
$= 2 \cos^{2} θ + \cos θ - 1 = 0$
$\Rightarrow$
$\cos 2 ϕ = \cos θ = - \frac{1}{4} \pm \sqrt{\frac{1}{16} + \frac{1}{2}}$
$\cos 2 ϕ = \cos θ = \frac{1}{2}$
$\Rightarrow ϕ = 30 ° N o w f r o m$
${(e^{2} p^{2})}_{m i n} = 2 (g \sin 2 ϕ) R (1 + \cos 2 ϕ)$
$\Rightarrow p_{m i n}^{2} = (\frac{2}{e^{2}}) (\frac{g \sqrt{3}}{2}) (\frac{3 R}{2})$
$p_{m i n} = \sqrt{\frac{3 \sqrt{3} g R}{2 e^{2}}}$
$u_{y, m i n}^{2} = p_{m i n}^{2} - 2 (g \sin 2 ϕ) R (1 + \cos 2 ϕ)$
$= \frac{3 \sqrt{3} g R}{2 e^{2}} - \frac{3 \sqrt{3} g R}{2}$
$u_{y, m i n} = \sqrt{\frac{3 \sqrt{3} g R (1 - e^{2})}{2 e^{2}}}$
$R \sin 2 ϕ = u_{x} t_{1} - (\frac{g \cos 2 ϕ}{2}) t_{1}^{2}$
$u_{x} = \frac{R \sqrt{3}}{2 t_{1}} + \frac{{g t}_{1}}{4}$
$t_{1} = \frac{p_{m i n} - u_{y, m i n}}{g \sin 2 ϕ} = 2 (\frac{p_{m i n} - u_{y, m i n}}{g \sqrt{3}})$
$t_{1} = \frac{2}{g \sqrt{3}} (\sqrt{\frac{3 \sqrt{3} g R}{2 e^{2}}}) (1 \pm \sqrt{1 - e^{2}})$
$u_{m i n}^{2} = (u_{x}^{2} + u_{y}^{2}) w i t h ϕ = 30 °$
$= {(\frac{R \sqrt{3}}{2 t_{1}} + \frac{{g t}_{1}}{4})}^{2} + \frac{3 \sqrt{3} g R (1 - e^{2})}{2 e^{2}}$

Commented byajfour last updated on 15/Feb/21

$P l e a s e r e v i e w i t S i r, i t d o n t$
$s e e m t o t a l y w i t h y o u r s o l u t i o n$
$a n d h e n c e w i t h t h e o r i g i n a l$
$q u e s t i o n, b u t y e t i t h i n k i t i s$
$a n i c e q u e s t i o n, a n d i t s a n s w e r$
$a s i h a v e d e r i v e d, d i d h a v e a$
$c h a n c e (i s u s p e c t e d) t o y i e l d$
$t h e s a m e a n s w e r \dots$

Commented bymr W last updated on 15/Feb/21

$l e t m e s o m e t i m e t o f o l l o w y o u r$
$w o r k i n g i n d e t a i l . i t h i n k y o u$
$i n t e r p r e t e d t h e q u e s t i o n d i f f e r e n t l y$
$a s i d i d . t h e q u e s t i o n s h o u l d a s k$
$w i t h w h i c h s m a l l e s t s p e e d o n e c a n$
$t h r o w t h e b a l l i n t o t h e b o w l s o t h a t$
$i t c a n r e t u r n b a c k t o t h e s t a r t p o i n t .$
$e . g . i f e = 0.75 t h e s m a l l e s t s p e e d i s$
$1.6082 \sqrt{g R} w h e n t h e b a l l f o l l o w s$
$f o l l o w i n g t r a c k :$

Commented bymr W last updated on 15/Feb/21

Commented bymr W last updated on 15/Feb/21

$f o r a n y o t h e r c a s e s, i t h i n k a l s o y o u r$
$r e s u l t, t h e b a l l n e e d s m o r e s p e e d$
$t h a n 1.6082 \sqrt{g R} .$
$i f e = 1, t h e s m a l l e s t s p e e d s h o u l d b e$
$0.9098 \sqrt{g R} .$

Commented bymr W last updated on 15/Feb/21

$a c c . t o y o u r r e s u l t, f o r e = 1,$
$t_{1} = \sqrt{\frac{2 \sqrt{3} R}{g}}$
$u_{m i n} = (\frac{\sqrt{3}}{2} + \frac{\sqrt{2 \sqrt{3}}}{4}) \sqrt{g R} \approx 1.3313 \sqrt{g R}$
$> 0.9098 \sqrt{g R} (w h a t i g o t)$

Commented bymr W last updated on 15/Feb/21

$i t h i n k t h e l e a s t e n e r g y l o s s d u r i n g$
$c o l l i s i o n {d o e s n}^{'} t m e a n t h e m i n i m u m$
$t o t a l e n e r g y t h e b a l l n e e d s .$

Commented byajfour last updated on 16/Feb/21

$I k n o w y o u a r e r i g h t S i r .$

Commented bymr W last updated on 16/Feb/21

$i t h i n k i h a v e f o u n d t h e l o g i c f a u l t$
$i n y o u r s o l u t i o n :$
$w i t h$
$B u t b a l l h a s t o r e a c h b a c k \Rightarrow$
${(e^{2} p^{2})}_{m i n} = 2 (g \sin 2 ϕ) R (1 + \cos 2 ϕ)$
$w e o n l y e n s u r e t h a t t h e b a l l c a n r e a c h$
$b a c k i t s h e i g h t (p o s i t i o n i n y -$
$d i r e c t i o n), t h e b a l l m a y n o t c o m e$
$b a c k t o i t s s t a r t p o i n t (a l s o i n x -$
$d i r e c t i o n), s o i t i s s o m e t h i n g l i k e$
$t h i s :$

Commented bymr W last updated on 16/Feb/21

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