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Question Number 109760 by ajfour last updated on 25/Aug/20

Commented by ajfour last updated on 25/Aug/20

$F i n d u i n t e r m s o f l, b, g s u c h t h a t$ $b o b h i t s h a n g i n g b l o c k .$
	Answered by mr W last updated on 25/Aug/20

	Commented by mr W last updated on 27/Aug/20
	$(1/2)mv^2 +mgl(1+sin θ)=(1/2)mu^2 ⇒v^2 =u^2 −2gl(1+sin θ) mg sin θ=((mv^2 )/l) ⇒u^2 =gl(2+3 sin θ) ⇒v^2 =gl sin θ R=l cos θ h=l(1−sin θ)−b t=(R/(v cos ϕ))=(l/(v tan θ)) h=v sin ϕ (l/(v tan θ))−(1/2)g ((l/(v tan θ)))^2 l(1−sin θ)−b=((l cos θ)/(tan θ))−((gl^2 )/(2v^2 tan^2 θ)) ((gl^2 )/(2v^2 tan^2 θ))=b+l((1/(sin θ))−1) v^2 =((gl^2 )/(2 tan^2 θ [b+l(((1−sin θ)/(sin θ)))])) gl sin θ=((gl^2 )/(2 tan^2 θ [b+l(((1−sin θ)/(sin θ)))])) sin θ=(1/(2 tan^2 θ ((b/l)−1+(1/(sin θ))))) 2 sin^3 θ ((b/l)−1+(1/(sin θ)))=1−sin^2 θ ⇒2(1−(b/l))sin^3 θ−3 sin^2 θ+1=0 ⇒(1/(sin θ))=2 sin [(1/3) sin^(−1) (1−(b/l))+((2π)/3)] ⇒u=(√({2+(3/(2 sin [(1/3) sin^(−1) (1−(b/l))+((2π)/3)]))}gl))$
	$\frac{1}{2} {m v}^{2} + m g l (1 + \sin θ) = \frac{1}{2} {m u}^{2}$ $\Rightarrow v^{2} = u^{2} - 2 g l (1 + \sin θ)$ $m g \sin θ = \frac{{m v}^{2}}{l}$ $\Rightarrow u^{2} = g l (2 + 3 \sin θ)$ $\Rightarrow v^{2} = g l \sin θ$ $R = l \cos θ$ $h = l (1 - \sin θ) - b$ $t = \frac{R}{v \cos φ} = \frac{l}{v \tan θ}$ $h = v \sin φ \frac{l}{v \tan θ} - \frac{1}{2} g {(\frac{l}{v \tan θ})}^{2}$ $l (1 - \sin θ) - b = \frac{l \cos θ}{\tan θ} - \frac{{g l}^{2}}{2 v^{2} \tan^{2} θ}$ $\frac{{g l}^{2}}{2 v^{2} \tan^{2} θ} = b + l (\frac{1}{\sin θ} - 1)$ $v^{2} = \frac{{g l}^{2}}{2 \tan^{2} θ [b + l (\frac{1 - \sin θ}{\sin θ})]}$ $g l \sin θ = \frac{{g l}^{2}}{2 \tan^{2} θ [b + l (\frac{1 - \sin θ}{\sin θ})]}$ $\sin θ = \frac{1}{2 \tan^{2} θ (\frac{b}{l} - 1 + \frac{1}{\sin θ})}$ $2 \sin^{3} θ (\frac{b}{l} - 1 + \frac{1}{\sin θ}) = 1 - \sin^{2} θ$ $\Rightarrow 2 (1 - \frac{b}{l}) \sin^{3} θ - 3 \sin^{2} θ + 1 = 0$ $\Rightarrow \frac{1}{\sin θ} = 2 \sin [\frac{1}{3} \sin^{- 1} (1 - \frac{b}{l}) + \frac{2 π}{3}]$ $\Rightarrow u = \sqrt{{2 + \frac{3}{2 \sin [\frac{1}{3} \sin^{- 1} (1 - \frac{b}{l}) + \frac{2 π}{3}]}} g l}$
	Commented by ajfour last updated on 27/Aug/20

	$l (1 - \sin θ) - b = \frac{v^{2} \sin^{2} φ}{2 g}$ $h = 1 - \sin θ - \frac{b}{l} = \frac{\sin θ \cos^{2} θ}{2}$ $\sin θ (2 + \cos^{2} θ) = 2 (1 - \frac{b}{l})$ $\sin θ (3 - \sin^{2} θ) = 2 (1 - \frac{b}{l})$ $. .$ $S i r i t h i n k e q . f o r \sin θ s h o u l d b e$ $t h i s . . .$ $. . . .$
	Commented by ajfour last updated on 27/Aug/20

	$m r W s i r, d o n t y o u t h i n k e v e n t h i s$ $e q . i s c o r r e c t . . ?$
	Commented by mr W last updated on 27/Aug/20

	$i n l i n e 1 y o u a s s u m e d t h a t t h e b a l l$ $r e a c h e s t h e h i g h e s t p o i n t w h e n i t h i t s$ $t h e b l o c k, i t h i n k t h i s i s n o t c o r r e c t .$
	Commented by mr W last updated on 27/Aug/20

	Commented by ajfour last updated on 27/Aug/20

	$t h a n k s s i r, c o n f u s i o n c l e a r .$
	Commented by mr W last updated on 27/Aug/20
	$u=(√({2+(3/(2 sin [(1/3) sin^(−1) (1−(b/l))+((2π)/3)]))}gl))$
	$u = \sqrt{{2 + \frac{3}{2 \sin [\frac{1}{3} \sin^{- 1} (1 - \frac{b}{l}) + \frac{2 π}{3}]}} g l}$
	Commented by mr W last updated on 29/Aug/20

	Commented by ajfour last updated on 29/Aug/20

	$n i c e o b s e r v a t i o n, S i r .$
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