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Question Number 72563 by ajfour last updated on 30/Oct/19

Commented by ajfour last updated on 30/Oct/19

$F i n d u_{m i n} f o r t h e s t o n e t o h i t$ $t h e s p h e r e a n y w h e r e .$
	Answered by mr W last updated on 30/Oct/19

	Commented by mr W last updated on 30/Oct/19

	$x = u \cos θ t$ $y = u \sin θ t - \frac{{g t}^{2}}{2}$ $y = x \tan θ - \frac{g (1 + \tan^{2} θ) x^{2}}{2 u^{2}}$ $w i t h t = \tan θ$ $\Rightarrow y = t x - \frac{g (1 + t^{2}) x^{2}}{2 u^{2}}$ $y^{'} = t - \frac{g (1 + t^{2}) x}{u^{2}}$ $a t p o i n t B (d - R \sin φ, R (1 - \cos φ)) :$ $R (1 - \cos φ) = t (d - R \sin φ) - \frac{g (1 + t^{2}) {(d - R \sin φ)}^{2}}{2 u^{2}} . . . (i)$ $- \tan φ = t - \frac{g (1 + t^{2}) (d - R \sin φ)}{u^{2}} . . . (i i)$ $\Rightarrow \frac{g (1 + t^{2}) (d - R \sin φ)}{u^{2}} = t + \tan φ$ $R (1 - \cos φ) = t (d - R \sin φ) - \frac{(d - R \sin φ) (t + \tan φ)}{2}$ $w i t h δ = \frac{d}{R}$ $\Rightarrow t = \tan φ + \frac{2 (1 - \cos φ)}{δ - \sin φ}$ $\Rightarrow \frac{g R}{u^{2}} = \frac{t + \tan φ}{(δ - \sin φ) (1 + t^{2})}$ $l e t U = \frac{g R}{2 u^{2}}$ $\Rightarrow 2 U = \frac{t + \tan φ}{(δ - \sin φ) (1 + t^{2})}$ $\Rightarrow 2 U = \frac{2 \tan φ + \frac{2 (1 - \cos φ)}{δ - \sin φ}}{(δ - \sin φ) [1 + {(\tan φ + \frac{2 (1 - \cos φ)}{δ - \sin φ})}^{2}]}$ $\Rightarrow U = \frac{1 - \cos φ + (δ - \sin φ) \tan φ}{{(δ - \sin φ)}^{2} + {[(δ - \sin φ) \tan φ + 2 (1 - \cos φ)]}^{2}}$ $\frac{d U}{d φ} = 0 \Rightarrow φ \Rightarrow U_{m a x} \Rightarrow u_{m i n}$ $e x a m p l e :$ $δ = \frac{d}{R} = \frac{6}{2} = 3$ $\Rightarrow φ = 0.7309$ $\Rightarrow U_{m a x} = 0.1922 \Rightarrow u_{m i n} = 1.613 \sqrt{g R}$
	Commented by mr W last updated on 30/Oct/19

	Commented by ajfour last updated on 30/Oct/19

	$T h a n k s S i r, b u t l e t s s e e i f a$ $f i n a l a n s w e r b e p o s s i b l e . .$
	Commented by malwaan last updated on 30/Oct/19

	$I t h i n k$ $u_{m a x} a t θ = 45^{°}$
	Commented by mr W last updated on 30/Oct/19

	$i t i s t o f i n d a p a r a b o l a w h i c h t a n g e n t s$ $t h e c i r c l e A N D h a s t h e m i n i m u m$ $e n e r g y . θ i s n o t a c o n s t a n t, b u t$ $d e p e n d s o n t h e p o s i t i o n o f t h e c i r c l e .$
	Commented by mr W last updated on 30/Oct/19

	Commented by mr W last updated on 30/Oct/19

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