Question-82057

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 = \frac{h - H}{a + b} \dots . . (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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