Question-165357

Question Number 165357 by ajfour last updated on 31/Jan/22

Commented by ajfour last updated on 31/Jan/22

W i t h w h a t u_{m i n} i s t h i s p o s s i b l e ?

Commented by ajfour last updated on 01/Feb/22

t h a n k s s i r, i l l t r y t o a r r i v e . .

Answered by ajfour last updated on 01/Feb/22

P a r a b o l a : y = a^{2} - x^{2}

l e t a t f i r s r c o l l i s i o n p o i n t

P (b, a^{2} - b^{2})

\frac{d y}{d x} = - 2 x = - 2 b = - \tan θ

\Rightarrow \tan θ = 2 b

t_{1} = \frac{2 a - b}{u_{x}}

a^{2} - b^{2} = u_{y} t_{1} - \frac{{g t}_{1}^{2}}{2}

\Rightarrow a^{2} - b^{2} = (2 a - b) \tan ϕ - \frac{g {(2 a - b)}^{2} (1 + \tan^{2} ϕ)}{2 u^{2}}

l e t j u s t b e f o r e r e a c h i n g P

v_{x} = u_{x}, v_{y} = u_{y} - {g t}_{1} \Rightarrow

\tan ϕ = (\frac{a^{2} - b^{2}}{2 a - b}) + \frac{g (2 a - b) (1 + \tan^{2} ϕ)}{2 u^{2}}

\dots . (i i)

o r \frac{2 u^{2}}{g (1 + 4 b^{2})} = \frac{2 a - b}{(\tan ϕ - \frac{a^{2} - b^{2}}{2 a - b})} . . (I)

\Rightarrow \frac{v_{y}}{v_{x}} = \tan ϕ = A + \frac{B}{u_{x}^{2}} \dots (i)

&

n o w d u e t o r e f l e c t i o n o f v e l o c i t y,

l e t j u s t a f t e r h i t t i n g P,

v e l o c i t y b e V .

V_{x} \cos θ + V_{y} \sin θ = u_{x} \cos θ + v_{y} \sin θ

&

V_{y} \cos θ - V_{x} \sin θ = u_{x} \sin θ - v_{y} \cos θ

\Rightarrow V_{x} + 2 {b V}_{y} = u_{x} + 2 {b v}_{y} &

V_{y} - 2 {b V}_{x} = 2 {b u}_{x} - v_{y}

\Rightarrow V_{y} = \frac{4 {b u}_{x} + (4 b^{2} - 1) v_{y}}{4 b^{2} + 1}

V_{x} = \frac{4 {b v}_{y} - (4 b^{2} - 1) u_{x}}{4 b^{2} + 1}

t_{2} = \frac{b}{V_{x}} & b^{2} = V_{y} t_{2} - \frac{{g t}_{2}^{2}}{2}

\Rightarrow b^{2} = b (\frac{V_{y}}{V_{x}}) - \frac{g}{2} {(\frac{b}{V_{x}})}^{2}

\Rightarrow b^{2} {\frac{4 {b v}_{y} - (4 b^{2} - 1) u_{x}}{4 b^{2} + 1}}^{2}

= b {\frac{4 {b v}_{y} - (4 b^{2} - 1) u_{x}}{4 b^{2} + 1}} {\frac{4 {b u}_{x} + (4 b^{2} - 1) v_{y}}{4 b^{2} + 1}}

- \frac{b^{2} g}{2}

u s i n g . . (i) & u_{x} = u \cos ϕ, u_{y} = u \sin ϕ

b^{2} {\frac{4 b \tan ϕ - (4 b^{2} - 1)}{4 b^{2} + 1}}^{2}

= b {\frac{4 b \tan ϕ - (4 b^{2} - 1)}{4 b^{2} + 1}} {\frac{4 b + (4 b^{2} - 1) \tan ϕ}{4 b^{2} + 1}}

- \frac{b^{2} g (1 + \tan^{2} ϕ)}{2 u^{2}}

\dots . (i i i)

u c a n b e o b t a i n e d i n t e r m s o f

\tan ϕ .

b y t h e w a y, i t h i n k f o r u_{m i n}

V_{y} = {g t}_{2} = \frac{4 {b u}_{x} + (4 b^{2} - 1) v_{y}}{4 b^{2} + 1}

\Rightarrow \frac{b g}{{\frac{4 {b v}_{y} - (4 b^{2} - 1) u_{x}}{4 b^{2} + 1}}} = \frac{4 {b u}_{x} + (4 b^{2} - 1) v_{y}}{4 b^{2} + 1}

\Rightarrow b g {(4 b^{2} + 1)}^{2} = u^{2} {4 b \tan ϕ - (4 b^{2} - 1)} {4 b + (4 b^{2} - 1) \tan ϕ}

& \frac{2 u^{2}}{(1 + 4 b^{2}) g} = \frac{2 a - b}{(\tan ϕ - \frac{a^{2} - b^{2}}{2 a - b})}

\Rightarrow \frac{2 b (1 + 4 b^{2}) (\tan ϕ - \frac{a^{2} - b^{2}}{2 a - b})}{2 a - b}

= {4 b \tan ϕ - (4 b^{2} - 1)} {4 b + (4 b^{2} - 1) \tan ϕ}

\dots . h o p e l e s s \dots .

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