Question-72672

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 - {A x}^{2} x^{2} + {(y - R)}^{2} = R^{2}

y = \frac{u^{2} \sin^{2} θ}{2 g} - {A x}^{2} \dots . (i)

\frac{2 u^{2} \sin θ \cos θ}{g} = 2 d \dots (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 \dots (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}}

\dots . .

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 \dots . .

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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