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Question Number 82057 by ajfour last updated on 17/Feb/20

Commented by ajfour last updated on 17/Feb/20

$T h e p l a n e d i v i n g a t s p e e d u d r o p s$ $a b a l l w h i c h u p o n o n e b o u n c e$ $c o l l i d e s w i t h t h e p l a n e w h e n$ $i t r e a c h e s i t s m a x i m u m h e i g h t .$ $F i n d t h e d e c l i n a t i o n o f \bar{u} f r o m$ $h o r i z o n t a l .$
	Answered by mr W last updated on 18/Feb/20

	Commented by mr W last updated on 18/Feb/20

	$e \sqrt{2 g h + {(u \sin θ)}^{2}} = \sqrt{2 {g h}_{2}}$ $\Rightarrow h_{2} = e^{2} (1 + \frac{u^{2} \sin^{2} θ}{2 g h}) h$ $t = \frac{\sqrt{u^{2} \sin^{2} θ + 2 g h} - u \sin θ + \sqrt{2 {g h}_{2}}}{g}$ $t = \frac{L}{u \cos θ} = \frac{h_{1}}{\tan θ u \cos θ} = \frac{h - h_{2}}{u \sin θ}$ $\Rightarrow L = \frac{1}{\tan θ} (1 - e^{2} - \frac{e^{2} u^{2} \sin^{2} θ}{2 g h}) h$ $\frac{\sqrt{u^{2} \sin^{2} θ + 2 g h} - u \sin θ + \sqrt{2 {g h}_{2}}}{g} = \frac{h}{u \sin θ} - \frac{h_{2}}{u \sin θ}$ $\sqrt{u^{2} \sin^{2} θ + 2 g h} - u \sin θ + \sqrt{2 {g h e}^{2} (1 + \frac{u^{2} \sin^{2} θ}{2 g h})} = \frac{g h}{u \sin θ} - \frac{g h}{u \sin θ} e^{2} (1 + \frac{u^{2} \sin^{2} θ}{2 g h})$ $(1 + e) \sqrt{u^{2} \sin^{2} θ + 2 g h} = \frac{(1 - e^{2}) g h}{u \sin θ} + \frac{(2 - e^{2}) u \sin θ}{2}$ $2 (1 + e) \sqrt{1 + \frac{2 g h}{u^{2} \sin^{2} θ}} = \frac{(1 - e^{2}) 2 g h}{u^{2} \sin^{2} θ} + 2 - e^{2}$ $l e t λ = \frac{2 g h}{u^{2} \sin^{2} θ}$ $2 (1 + e) \sqrt{1 + λ} = (1 - e^{2}) λ + 2 - e^{2}$ ${(1 - e^{2})}^{2} λ^{2} - 2 e (1 + e) (4 + e - e^{2}) λ - e (8 + 7 e - e^{3}) = 0$ $λ = \frac{e (4 + e - e^{2}) + \sqrt{e (8 + 7 e + 2 e^{2} - e^{3})}}{(1 + e) {(1 - e)}^{2}}$ $\Rightarrow \frac{2 g h}{u^{2} \sin^{2} θ} = \frac{e (4 + e - e^{2}) + \sqrt{e (8 + 7 e + 2 e^{2} - e^{3})}}{(1 + e) {(1 - e)}^{2}}$ $\Rightarrow θ = \sin^{- 1} [\frac{(1 - e) \sqrt{1 + e}}{\sqrt{e (4 + e - e^{2}) + \sqrt{e (8 + 7 e + 2 e^{2} - e^{3})}}} \sqrt{\frac{2 g h}{u^{2}}}]$ $e x a m p l e : e = 0.5$ $θ \approx \sin^{- 1} (0.286716 \sqrt{\frac{2 g h}{u^{2}}})$
	Commented by ajfour last updated on 18/Feb/20

	$V e r y N i c e, S i r . A w f u l l y G o o d!$
	Commented by mr W last updated on 18/Feb/20

	Commented by mr W last updated on 18/Feb/20

	Commented by mr W last updated on 18/Feb/20

	Answered by ajfour last updated on 18/Feb/20
	$tan θ=((h−H)/(a+b)) .....(i) e^2 v_y ^2 =e^2 (u^2 sin^2 θ+2gh) H=((e^2 v_y ^2 )/(2g))=((e^2 (u^2 sin^2 θ+2gh))/(2g)) (usin θ)t_1 +((gt_1 ^2 )/2)=h ⇒ t_1 =−((usin θ)/g)+(√(((u^2 sin^2 θ)/g^2 )+((2h)/g))) & t_2 =((ev_y )/g)=e(√(((u^2 sin^2 θ)/g^2 )+((2h)/g))) a+b=(ucos θ)(t_1 +t_2 ) from ..(i) (tan θ)(a+b)=h−H usin θ{−((usin θ)/g)+(√(((u^2 sin^2 θ)/g^2 )+((2h)/g))) +e(√(((u^2 sin^2 θ)/g^2 )+((2h)/g))) } = h−((e^2 (u^2 sin^2 θ+2gh))/(2g)) let ((usin θ)/g)=x, ((2h)/g)=c ⇒ x{−x+(√(x^2 +c))+e(√(x^2 +c)) } = (c/2)−(e^2 /2)(x^2 +c) ⇒ {x^2 (1−(e^2 /2))+(c/2)(1−e^2 )}^2 = x^2 (1+e)^2 (x^2 +c) For e=1/2 , also let x^2 /c = z {(7/8)z+(3/8)}^2 =(9/4)z(z+1) 49z^2 +42z+9=144z^2 +144z 95z^2 +102z−9=0 z=−((51)/(95))+(√((((51)/(95)))^2 +((9/(95))))) ≈0.08197636 ⇒ (((usin θ)/g))^2 =((2h)/g)×0.08197636 sin θ=((√(2gh))/u)×0.286315$
	$\tan θ = \frac{h - H}{a + b} . . . . . (i)$ $e^{2} v_{y}^{2} = e^{2} (u^{2} \sin^{2} θ + 2 g h)$ $H = \frac{e^{2} v_{y}^{2}}{2 g} = \frac{e^{2} (u^{2} \sin^{2} θ + 2 g h)}{2 g}$ $(u \sin θ) t_{1} + \frac{{g t}_{1}^{2}}{2} = h$ $\Rightarrow t_{1} = - \frac{u \sin θ}{g} + \sqrt{\frac{u^{2} \sin^{2} θ}{g^{2}} + \frac{2 h}{g}}$ $& t_{2} = \frac{{e v}_{y}}{g} = e \sqrt{\frac{u^{2} \sin^{2} θ}{g^{2}} + \frac{2 h}{g}}$ $a + b = (u \cos θ) (t_{1} + t_{2})$ $f r o m . . (i)$ $(\tan θ) (a + b) = h - H$ $u \sin θ {- \frac{u \sin θ}{g} + \sqrt{\frac{u^{2} \sin^{2} θ}{g^{2}} + \frac{2 h}{g}}$ $+ e \sqrt{\frac{u^{2} \sin^{2} θ}{g^{2}} + \frac{2 h}{g}}}$ $= h - \frac{e^{2} (u^{2} \sin^{2} θ + 2 g h)}{2 g}$ $l e t \frac{u \sin θ}{g} = x, \frac{2 h}{g} = c \Rightarrow$ $x {- x + \sqrt{x^{2} + c} + e \sqrt{x^{2} + c}}$ $= \frac{c}{2} - \frac{e^{2}}{2} (x^{2} + c)$ $\Rightarrow$ ${x^{2} (1 - \frac{e^{2}}{2}) + \frac{c}{2} (1 - e^{2})}^{2}$ $= x^{2} {(1 + e)}^{2} (x^{2} + c)$ $F o r e = 1 / 2, a l s o l e t x^{2} / c = z$ ${\frac{7}{8} z + \frac{3}{8}}^{2} = \frac{9}{4} z (z + 1)$ $49 z^{2} + 42 z + 9 = 144 z^{2} + 144 z$ $95 z^{2} + 102 z - 9 = 0$ $z = - \frac{51}{95} + \sqrt{{(\frac{51}{95})}^{2} + (\frac{9}{95})}$ $\approx 0.08197636$ $\Rightarrow {(\frac{u \sin θ}{g})}^{2} = \frac{2 h}{g} \times 0.08197636$ $\sin θ = \frac{\sqrt{2 g h}}{u} \times 0.286315$
	Commented by ajfour last updated on 18/Feb/20

	$Yes sir, i agree, i shall reexamine$ $this one later, i intend to try to$ $obtain s_{\max}, s_{\min} of our eql . △ Q$ $presently . .$
	Commented by mr W last updated on 18/Feb/20

	$o u r r e s u l t s d i f f e r f r o m e a c h o t h e r .$ $i {c a n}^{'} t s a y w h e r e m y o r y o u r e r r o r i s .$ $i g r a p h e d m y r e s u l t f o r v a r i o u s c a s e s,$ $b o t h c u r v e s m e e t e x a c t l y o n a p o i n t$ $o n t h e x - a x i s, s o m y r e s u l t s e e m s t o$ $b e o k .$
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