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Question Number 72672 by mr W last updated on 31/Oct/19

Commented by mr W last updated on 31/Oct/19

$f i n d u_{m i n} f o r t h e s t o n e t o h i t t h e$ $g r o u n d b e h i n d t h e s p h e r e .$ $(s e e a l s o Q 72563, w i t h t h a n k s t o$ $a j f o u r s i r f o r t h e o r g i n a l q u e s t i o n)$
Commented by malwaan last updated on 31/Oct/19

$v e r y e a s y!$
	Answered by ajfour last updated on 31/Oct/19
	$y=H−Ax^2 x^2 +(y−R)^2 =R^2 y=((u^2 sin^2 θ)/(2g))−Ax^2 ....(i) ((2u^2 sin θcos θ)/g)=2d ...(ii) x^2 +(((u^2 sin^2 θ)/(2g))−Ax^2 −R)^2 =R^2 ⇒ x^2 −2A(((u^2 sin^2 θ)/(2g))−R)x^2 +A^2 x^4 +(((u^2 sin^2 θ)/(2g))−R)^2 −R^2 =0 coeff. of x^2 =0 { ((further if ((u^2 sin^2 θ)/(2g))=2R)),((stone grazes the sphere top)) :} ⇒ 2A(((u^2 sin^2 θ)/(2g))−R)=1 ...(iii) for x=d , y=0 eq. (i) then gives Ad^2 =((u^2 sin^2 θ)/(2g)) Now using (iii) in here d^2 =((u^2 sin^2 θ)/(2g))×2(((u^2 sin^2 θ)/(2g))−R) And from (ii) u^2 =((gd)/(sin θcos θ)) ⇒ ((u^2 sin^2 θ)/(2g))=((dtan θ)/2) ⇒ d^2 = dtan θ(((dtan θ)/2)−R) t= tan θ dt^2 −2Rt−2d=0 ⇒ t=tan θ=((2R+(√(4R^2 +8d^2 )))/(2d)) θ=tan^(−1) ((R/d)+(√(((R/d))^2 +2)) ) u^2 =((gdtan θ)/(1+tan^2 θ)) u^2 =((gd(((2R+(√(4R^2 +8d^2 )))/(2d))))/(1+(((2R+(√(4R^2 +8d^2 )))/(2d)))^2 )) .....$
	$y = H - {A x}^{2} x^{2} + {(y - R)}^{2} = R^{2}$ $y = \frac{u^{2} \sin^{2} θ}{2 g} - {A x}^{2} . . . . (i)$ $\frac{2 u^{2} \sin θ \cos θ}{g} = 2 d . . . (i i)$ $x^{2} + {(\frac{u^{2} \sin^{2} θ}{2 g} - {A x}^{2} - R)}^{2} = R^{2}$ $\Rightarrow$ $x^{2} - 2 A (\frac{u^{2} \sin^{2} θ}{2 g} - R) x^{2} + A^{2} x^{4}$ $+ {(\frac{u^{2} \sin^{2} θ}{2 g} - R)}^{2} - R^{2} = 0$ $c o e f f . o f x^{2} = 0$ ${\begin{cases} f u r t h e r i f \frac{u^{2} \sin^{2} θ}{2 g} = 2 R \\ s t o n e g r a z e s t h e s p h e r e t o p \end{cases}$ $\Rightarrow 2 A (\frac{u^{2} \sin^{2} θ}{2 g} - R) = 1 . . . (i i i)$ $f o r x = d, y = 0 e q . (i) t h e n g i v e s$ ${A d}^{2} = \frac{u^{2} \sin^{2} θ}{2 g}$ $N o w u s i n g (i i i) i n h e r e$ $d^{2} = \frac{u^{2} \sin^{2} θ}{2 g} \times 2 (\frac{u^{2} \sin^{2} θ}{2 g} - R)$ $A n d f r o m (i i)$ $u^{2} = \frac{g d}{\sin θ \cos θ} \Rightarrow \frac{u^{2} \sin^{2} θ}{2 g} = \frac{d \tan θ}{2}$ $\Rightarrow d^{2} = d \tan θ (\frac{d \tan θ}{2} - R)$ $t = \tan θ$ ${d t}^{2} - 2 R t - 2 d = 0$ $\Rightarrow t = \tan θ = \frac{2 R + \sqrt{4 R^{2} + 8 d^{2}}}{2 d}$ $θ = \tan^{- 1} (\frac{R}{d} + \sqrt{{(\frac{R}{d})}^{2} + 2})$ $u^{2} = \frac{g d \tan θ}{1 + \tan^{2} θ}$ $u^{2} = \frac{g d (\frac{2 R + \sqrt{4 R^{2} + 8 d^{2}}}{2 d})}{1 + {(\frac{2 R + \sqrt{4 R^{2} + 8 d^{2}}}{2 d})}^{2}}$ $. . . . .$
	Answered by mr W last updated on 31/Oct/19

	Commented by mr W last updated on 31/Oct/19

	$t h e r e a r e m a n y d i f f e r e n t c a s e s . w h a t$ $t h e i m a g e s h o w s i s t h e s i t u a t i o n w h e n$ $t h e s p h e r e i s f a r a w a y . i f t h e s p h e r e$ $i s n e a r f r o m t h e o r i g i n, i t c o u l d l o o k$ $d i f f e r e n t l y . m y m e t h o d c a n n o t$ $t r e a t s u c h c a s e s c o r r e c t l y . y o u r s o l u t i o n$ $i s f o r s u c h c a s e s t h e n .$
	Commented by mr W last updated on 31/Oct/19

	$x = u \cos θ t$ $y = u \sin θ t - \frac{{g t}^{2}}{2}$ $y = x \tan θ - \frac{g}{2 u^{2}} (1 + \tan^{2} θ) x^{2}$ $l e t t = \tan θ$ $y = t x - \frac{g}{2 u^{2}} (1 + t^{2}) x^{2}$ $y^{'} = t - \frac{g (1 + t^{2}) x}{u^{2}}$ $B (d + R \sin φ, R (1 + \cos φ))$ $- \tan φ = t - \frac{g (1 + t^{2}) (d + R \sin φ)}{u^{2}}$ $\Rightarrow \frac{g (1 + t^{2}) (d + R \sin φ)}{u^{2}} = t + \tan φ$ $R (1 + \cos φ) = t (d + R \sin φ) - \frac{g}{2 u^{2}} (1 + t^{2}) {(d + R \sin φ)}^{2}$ $R (1 + \cos φ) = t (d + R \sin φ) - \frac{(d + R \sin φ) (t + \tan φ)}{2}$ $\Rightarrow t = \tan φ + \frac{2 (1 + \cos φ)}{δ + \sin φ} w i t h δ = \frac{d}{R}$ $\Rightarrow \frac{g R (1 + t^{2}) (δ + \sin φ)}{u^{2}} = t + \tan φ$ $\Rightarrow \frac{g R}{u^{2}} = \frac{t + \tan φ}{(δ + \sin φ) (1 + t^{2})}$ $\Rightarrow \frac{g R}{2 u^{2}} = U = \frac{1 + \cos φ + (δ + \sin φ) \tan φ}{{(δ + \sin φ)}^{2} + {[(δ + \sin φ) \tan φ + 2 (1 + \cos φ)]}^{2}}$ $\frac{d U}{d φ} = 0 . . . . .$
	Commented by mr W last updated on 31/Oct/19

	Commented by ajfour last updated on 31/Oct/19

	$s h o u l d i t b e t h i s w a y, i w a s$ $m i s t a k e n t h e n s i r, i t h o u g h t$ $t h e t r a j e c t o r y w o u l d b e e n v e l o p$ $t h e s p h e r e s y m m e t r i c a l l y .$
	Commented by mr W last updated on 31/Oct/19

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