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Question Number 42654 by ajfour last updated on 30/Aug/18

Commented by ajfour last updated on 30/Aug/18

$I f t h e r e d b a l l i s t o h i t t h e w a l l$ $a n d t h e n t h e b l u e b a l l, f i n d t h e$ $s p e e d u a n d h e i g h t h w h e r e i t$ $h i t s t h e w a l l . (c o e f f i c i e n t o f$ $r e s t i t u t i o n b e i n g e) .$
Commented by MrW3 last updated on 03/Sep/18

$t h e q u e s t i o n c o u l d b e m o d i f i e d t o :$ $f i n d t h e s m a l l e s t s p e e d o f t h e r e d b a l l$ $s u c h t h a t i t c a n h i t t h e w a l l a n d t h e n$ $t h e b l u e b a l l .$
	Answered by MrW3 last updated on 04/Sep/18

	$a = u \cos θ t_{1} . . . (i)$ $h - p = u \sin θ t_{1} - \frac{1}{2} g t_{1}^{2} . . . (i i)$ $a + b = e u \cos θ t_{2} . . . (i i i)$ $q - h = (u \sin θ - g t_{1}) t_{2} - \frac{1}{2} g t_{2}^{2} . . . (i v)$ $w e c a n s o l v e t h i s o n l y i f θ o r h i s g i v e n .$ ${l e t}^{'} s s a y h i s g i v e n .$ $h - p = u \sin θ \frac{a}{u \cos θ} - \frac{g}{2} {(\frac{a}{u \cos θ})}^{2}$ $h - p = a \tan θ - \frac{a^{2} g}{2 u^{2} \cos^{2} θ}$ $\Rightarrow h - p = a \tan θ - \frac{a^{2} g}{2 u^{2}} (1 + \tan^{2} θ) . . . (v)$ $q - h = (u \sin θ - \frac{a g}{u \cos θ}) \frac{a + b}{e u \cos θ} - \frac{g}{2} {(\frac{a + b}{e u \cos θ})}^{2}$ $q - h = \frac{a + b}{e} \tan θ - (a + b) (2 a + \frac{a + b}{e}) \frac{g}{2 {e u}^{2} \cos^{2} θ}$ $\Rightarrow q - h = \frac{a + b}{e} \tan θ - (a + b) (2 a + \frac{a + b}{e}) \frac{g}{2 {e u}^{2}} (1 + \tan^{2} θ) . . . (v i)$ $(i v) \times a^{2} :$ $a^{2} (q - h) = \frac{(a + b) a^{2}}{e} \tan θ - \frac{a + b}{e} (2 a + \frac{a + b}{e}) (a \tan θ + p - h)$ $e^{2} a^{2} (q - h) = {e a}^{2} (a + b) \tan θ - (a + b) (2 e a + a + b) (a \tan θ + p - h)$ $a (a + b) (e a + a + b) \tan θ = e^{2} a^{2} (h - q) + (a + b) (2 e a + a + b) (h - p)$ $\Rightarrow \tan θ = \frac{e^{2} a^{2} (h - q) + (a + b) (2 e a + a + b) (h - p)}{a (a + b) (e a + a + b)} = λ$ $f r o m (v) :$ $u = \sqrt{\frac{a^{2} (1 + \tan^{2} θ) g}{2 (a \tan θ - h + p)}}$ $\Rightarrow u = a \sqrt{\frac{(1 + λ^{2}) g}{2 (λ a - h + p)}}$ $a l t e r n a t i v e l y :$ $(v) + (v i) :$ $q - p = (a + \frac{a + b}{e}) \tan θ - [a^{2} + \frac{a + b}{e} (2 a + \frac{a + b}{e})] \frac{g}{2 u^{2}} (1 + \tan^{2} θ)$ $q - p = (a + \frac{a + b}{e}) \tan θ - {(a + \frac{a + b}{e})}^{2} \frac{g}{2 u^{2}} (1 + \tan^{2} θ)$ $w i t h A = a + \frac{a + b}{e}, B = q - p, λ = \tan θ$ $w e g e t$ $B = A λ - A^{2} (1 + λ^{2}) \frac{g}{2 u^{2}}$ $\Rightarrow u^{2} = \frac{{g A}^{2}}{2} \times \frac{1 + λ^{2}}{A λ - B}$ $l e t f (λ) = \frac{1 + λ^{2}}{A λ - B}$ $\frac{d f (λ)}{d λ} = \frac{2 λ}{A λ - B} - \frac{A (1 + λ^{2})}{{(A λ - B)}^{2}} = 0$ $\Rightarrow \frac{1 + λ^{2}}{A λ - B} = \frac{2 λ}{A}$ $\Rightarrow 2 λ (A λ - B) - A (1 + λ^{2}) = 0$ $\Rightarrow A λ^{2} - 2 B λ - A = 0$ $\Rightarrow λ = \frac{B + \sqrt{A^{2} + B^{2}}}{A} = \frac{q - p}{a + \frac{a + b}{e}} + \sqrt{1 + {(\frac{q - p}{a + \frac{a + b}{e}})}^{2}}$ $\Rightarrow u_{m i n}^{2} = \frac{{g A}^{2}}{2} \times \frac{2 λ}{A} = g (B + \sqrt{A^{2} + B^{2}})$ $\Rightarrow u_{m i n} = \sqrt{[(q - p) + \sqrt{{(a + \frac{a + b}{e})}^{2} + {(q - p)}^{2}}] g}$ $h_{m} = p + a λ - \frac{a^{2} g}{2 u_{m i n}^{2}} (1 + λ^{2})$ $h_{m} = p + a λ - \frac{a^{2} (1 + λ^{2})}{2 A λ}$ $h_{m} = p + a λ - \frac{a^{2} (A λ - B)}{A^{2}}$ $h_{m} = p + a (1 - \frac{a}{A}) (\frac{B}{A} + \frac{\sqrt{A^{2} + B^{2}}}{A}) + \frac{a^{2} B}{A^{2}}$ $\Rightarrow h_{m} = p + a (1 - \frac{a}{a + \frac{a + b}{e}}) [\frac{q - p}{a + \frac{a + b}{e}} + \sqrt{1 + {(\frac{q - p}{a + \frac{a + b}{e}})}^{2}}] + {(\frac{a}{a + \frac{a + b}{e}})}^{2} (q - p)$ $θ_{m} = \tan^{- 1} λ$ $\Rightarrow θ_{m} = \tan^{- 1} [\frac{q - p}{a + \frac{a + b}{e}} + \sqrt{1 + {(\frac{q - p}{a + \frac{a + b}{e}})}^{2}}]$
	Commented by MrW3 last updated on 02/Sep/18

	${l e t}^{'} s a s s u m e t h e w a l l i s ‘ ‘ {p e r m e a b l e}^{″},$ $t h e b a l l c a n g o t h r o u g h i t i n s t e a d o f$ $i s r e f l e c t e d . {l e t}^{'} s c o n s i d e r e = 1 (s e e a b o v e) .$ $w e c a n s e e t h e b a l l h a s d i f f e r e n t w a y s t o$ $h i t t h e w a l l (a t p o i n t B) a n d t h e p o i n t C .$ $i . e . w e c a n n o t d e t e r m i n e b o t h h a n d u .$ $o n e o f t h e m m u s t b e g i v e n .$
	Commented by MrW3 last updated on 04/Sep/18

	$a n d w e c a n s e e t h e r e e x i s t s a p o i n t B$ $w i t h h = h_{m} a t w h i c h t h e b a l l c a n r e a c h$ $p o i n t C w i t h a m i n i m a l i n i t i a l s p e e d u_{m i n} .$
	Commented by MrW3 last updated on 02/Sep/18

	Commented by ajfour last updated on 02/Sep/18

	$t h a n k y o u S i r, f o r t h e e f f o r t f u l$ $r e s p o n s e; a n d c o r r e c t i n g t h e$ $q u e s t i o n e v e n . I s h a l l a l s o t r y t o$ $s o l v e a n d m a t c h m y a n s w e r w i t h$ $y o u r s .$
	Commented by MrW3 last updated on 03/Sep/18

	Commented by MrW3 last updated on 03/Sep/18

	$f o l l o w i n g e x a m p l e s h o w s t h e c a s e :$ $a = 3 m, b = 2 m, e = 0.5$ $p = 1 m, q = 2 m$ $w e s e e a t h = 3.55 m, u_{m i n} = 11.85 m / s$
	Commented by ajfour last updated on 03/Sep/18

	$A w e s o m e W o r k i n g S i r .$
	Answered by ajfour last updated on 04/Sep/18

	$S i n c e t h e v e r t i c a l c o m p o n e n t o f$ $u s u f f e r s n o c h a n g e u p o n h i t t i n g$ $t h e w a l l;$ $q - p = u t \sin θ - \frac{{g t}^{2}}{2} . . . (i)$ $a n d t = \frac{a}{u \cos θ} + \frac{a + b}{e u \cos θ} . . . (i i)$ $u s i n g (i i) i n (i) :$ $q - p = u \sin θ [\frac{a}{u \cos θ} + \frac{a + b}{e u \cos θ}]$ $- \frac{g}{2} {[\frac{a}{u \cos θ} + \frac{a + b}{e u \cos θ}]}^{2}$ $\Rightarrow q - p = \tan θ (a + \frac{a + b}{e})$ $- \frac{g}{2 u^{2}} (1 + \tan^{2} θ) {(a + \frac{a + b}{e})}^{2}$ $i f m = \tan θ; q - p = B; a + \frac{a + b}{e} = A$ $\frac{{g A}^{2}}{2 u^{2}} = \frac{m A - B}{1 + m^{2}} . . . . (i i i)$ $F o r m i n i m u m u$ $\frac{d (\frac{{g A}^{2}}{2 u^{2}})}{d m} = 0 \Rightarrow$ $A (1 + m^{2}) = 2 m (m A - B)$ $o r {A m}^{2} - 2 B m - A = 0$ $\Rightarrow m = \frac{B}{A} + \sqrt{1 + \frac{B^{2}}{A^{2}}}$ $u s i n g t h i s v a l u e i n e q . (i i i)$ $\frac{{g A}^{2}}{2 u_{m i n}^{2}} = \frac{\sqrt{A^{2} + B^{2}}}{1 + {(\frac{B}{A} + \sqrt{1 + \frac{B^{2}}{A^{2}}})}^{2}}$ $\Rightarrow u_{m i n}^{2} = \frac{g}{2} \times \frac{A^{2} + {(B + \sqrt{A^{2} + B^{2}})}^{2}}{\sqrt{A^{2} + B^{2}}}$ $u_{m i n}^{2} = g (\sqrt{A^{2} + B^{2}} + B)$ $= g [\sqrt{{(a + \frac{a + b}{e})}^{2} + {(q - p)}^{2}} + (q - p)]$ $h = p + a m - \frac{a^{2} g (1 + m^{2})}{2 u_{m i n}^{2}}$ $w i t h a = 3, b = 2, e = 0.5,$ $p = 1, q = 2, g = 10, w e h a v e$ $u_{m i n}^{2} = 10 [\sqrt{170} + 1]$ $u_{m i n} = \sqrt{140.38} \approx 11.85 m / s$ $m = \frac{1}{13} + \sqrt{1 + \frac{1}{169}} = \frac{1 + \sqrt{170}}{13} \approx 1.08$ $h = 1 + 3 \times 1.08 - \frac{45 \times 2.16}{140.38}$ $\approx 3.55 m .$
	Commented by ajfour last updated on 03/Sep/18

	$S i r, p l e a s e r e s o l v e t h e m i s m a t c h!$
	Commented by MrW3 last updated on 03/Sep/18

	$i t c a n b e s i m p l i f i e d t o :$ $u_{m i n}^{2} = g (B + \sqrt{A^{2} + B^{2}})$ $u_{m i n}^{2} = g [(q - p) + \sqrt{{(a + \frac{a + b}{e})}^{2} + {(q - p)}^{2}}]$ $= 10 \times (1 + \sqrt{170})$ $\Rightarrow u_{m i n} = 11.85 m / s$ $I^{'} l l c h e c k m y w o r k i n g l a t e r .$
	Commented by MrW3 last updated on 04/Sep/18

	$y o u r s o l u t i o n i s b e s t! s h o r t a n d c l e a r .$
	Commented by MrW3 last updated on 03/Sep/18

	$I f o u n d a m i s t a k e i n m y w o r k i n g . {i t}^{'} s$ $f i x e d . n o w w e h a v e t h e s a m e r e s u l t .$
	Commented by ajfour last updated on 04/Sep/18

	$t h a n k s S i r, c o r r e c t e d .$
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