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Question Number 197619 by mr W last updated on 24/Sep/23

Commented by mr W last updated on 24/Sep/23

$a s s u m e t h e h i l l h a s t h e s h a p e o f a$ $p a r a b o l a . f i n d t h e m i n i m u m s p e e d$ $w i t h w h i c h a p r o j e c t i l e s h o u l d b e$ $l a u n c h e d f r o m p o i n t C s u c h t h a t i t$ $c a n h i t p o i n t B .$
	Answered by mahdipoor last updated on 24/Sep/23
	$position of ball ⇒ { ((x=ut.cosβ)),((y=ut.sinβ−((gt^2 )/2)=x.tanβ−((gx^2 )/(2u^2 cos^2 β)))) :} I) when x=b+2a⇒y=0 0=(b+2a)[tanβ−((g(b+2a))/(2u^2 cos^2 β))]⇒(g/(2u^2 cos^2 β))=((tanβ)/(b+2a)) ⇒u=(√(((g(b+2a))/2)((1/(tanβ))+tanβ))) II) 1}h−y=(h/a^2 )(x−(a+b))^2 2}y=x.tanβ−((gx^2 )/(2u^2 cos^2 β))=x.tanβ[1−(x/(b+2a))] 1,2 ⇒(((tanβ)/(b+2a))−(h/a^2 ))x^2 +(((2h(a+b))/a^2 )−tanβ)x −((h(2ab+b^2 ))/a^2 )=0 ⇒((x.tanβ)/(b+2a))(x−(b+2a))−(h/a^2 )(x−(b+2a))(x−b)=0 ⇒(x−(b+2a))(x(((tanβ)/(b+2a))−(h/a^2 ))+((hb)/a^2 ))=0 ⇒^1 x_2 =((hb/a^2 )/(h/a^2 −tanβ/(b+2a)))≤0 or ∄⇒(h/a^2 )≤((tanβ)/(b+2a)) ⇒((h(b+2a))/a^2 )≤tanβ ⇒^2 x_2 =((hb/a^2 )/(h/a^2 −tanβ/(b+2a)))≥(b+2a)⇒((tanβ−2h/a)/(h/a^2 −tanβ/(b+2a)))≥0 ⇒ { ((((2h)/a)≤tanβ<((h(b+2a))/a^2 ))),((((2h)/a)≥tanβ>((h(b+2a))/a^2 ) impossible (2>2+(b/a)))) :} ⇒((2h)/a)≤tanβ<((2h(b+2a))/a^2 ) ⇒^(1,2) ((2h)/a)≤tanβ f=min((1/(tanβ))+tanβ) = { ((2 if ((2h)/a)≤1)),((((2h)/a)+(a/(2h)) if ((2h)/a)>1)) :} ⇒⇒⇒⇒min u=(√((g(b+2a)f)/2))$
	$p o s i t i o n o f b a l l \Rightarrow$ ${\begin{cases} x = u t . c o s β \\ y = u t . s i n β - \frac{{g t}^{2}}{2} = x . t a n β - \frac{{g x}^{2}}{2 u^{2} {c o s}^{2} β} \end{cases}$ $I) w h e n x = b + 2 a \Rightarrow y = 0$ $0 = (b + 2 a) [t a n β - \frac{g (b + 2 a)}{2 u^{2} {c o s}^{2} β}] \Rightarrow \frac{g}{2 u^{2} {c o s}^{2} β} = \frac{t a n β}{b + 2 a}$ $\Rightarrow u = \sqrt{\frac{g (b + 2 a)}{2} (\frac{1}{t a n β} + t a n β)}$ $II)$ $1} h - y = \frac{h}{a^{2}} {(x - (a + b))}^{2}$ $2} y = x . t a n β - \frac{{g x}^{2}}{2 u^{2} {c o s}^{2} β} = x . t a n β [1 - \frac{x}{b + 2 a}]$ $1, 2 \Rightarrow (\frac{t a n β}{b + 2 a} - \frac{h}{a^{2}}) x^{2} + (\frac{2 h (a + b)}{a^{2}} - t a n β) x$ $- \frac{h (2 a b + b^{2})}{a^{2}} = 0$ $\Rightarrow \frac{x . t a n β}{b + 2 a} (x - (b + 2 a)) - \frac{h}{a^{2}} (x - (b + 2 a)) (x - b) = 0$ $\Rightarrow (x - (b + 2 a)) (x (\frac{t a n β}{b + 2 a} - \frac{h}{a^{2}}) + \frac{h b}{a^{2}}) = 0$ $\overset{1}{\Rightarrow} x_{2} = \frac{h b / a^{2}}{h / a^{2} - t a n β / (b + 2 a)} ⩽ 0 o r ∄ \Rightarrow \frac{h}{a^{2}} ⩽ \frac{t a n β}{b + 2 a}$ $\Rightarrow \frac{h (b + 2 a)}{a^{2}} ⩽ t a n β$ $\overset{2}{\Rightarrow} x_{2} = \frac{h b / a^{2}}{h / a^{2} - t a n β / (b + 2 a)} ⩾ (b + 2 a) \Rightarrow \frac{t a n β - 2 h / a}{h / a^{2} - t a n β / (b + 2 a)} ⩾ 0$ $\Rightarrow {\begin{cases} \frac{2 h}{a} ⩽ t a n β < \frac{h (b + 2 a)}{a^{2}} \\ \frac{2 h}{a} ⩾ t a n β > \frac{h (b + 2 a)}{a^{2}} i m p o s s i b l e (2 > 2 + \frac{b}{a}) \end{cases}$ $\Rightarrow \frac{2 h}{a} ⩽ t a n β < \frac{2 h (b + 2 a)}{a^{2}}$ $\overset{1, 2}{\Rightarrow} \frac{2 h}{a} ⩽ t a n β$ $f = m i n (\frac{1}{t a n β} + t a n β) = {\begin{cases} 2 i f \frac{2 h}{a} ⩽ 1 \\ \frac{2 h}{a} + \frac{a}{2 h} i f \frac{2 h}{a} > 1 \end{cases}$ $\Rightarrow\Rightarrow\Rightarrow\Rightarrow m i n u = \sqrt{\frac{g (b + 2 a) f}{2}}$
	Commented by mr W last updated on 24/Sep/23

	$t h a n k s s i r!$
	Answered by mr W last updated on 24/Sep/23

	Commented by mr W last updated on 25/Sep/23

	$x = u \cos θ t = 2 a + b$ $\Rightarrow t = \frac{2 a + b}{u \cos θ}$ $y = u \sin θ t - \frac{{g t}^{2}}{2} = 0$ $u \sin θ - \frac{g}{2} \times \frac{2 a + b}{u \cos θ} = 0$ $\Rightarrow u^{2} = \frac{(2 a + b) g}{\sin 2 θ}$ $ϕ = \tan^{- 1} \frac{2 h}{a}$ $θ ⩾ ϕ = \tan^{- 1} \frac{2 h}{a}$ $i f ϕ ⩽ \frac{π}{4} = 45 °, i . e . \frac{h}{a} ⩽ \frac{1}{2} :$ $u_{m i n}^{2} = (2 a + b) g$ $\Rightarrow u_{m i n} = \sqrt{(2 a + b) g}$ $i f ϕ > \frac{π}{4} = 45 °, i . e . \frac{h}{a} > \frac{1}{2} :$ $u_{m i n}^{2} = \frac{(2 a + b) g}{\sin 2 ϕ} = (\frac{a}{4 h} + \frac{h}{a}) (2 a + b) g$ $\Rightarrow u_{m i n} = \sqrt{(\frac{a}{4 h} + \frac{h}{a}) (2 a + b) g}$
	Commented by mahdipoor last updated on 25/Sep/23

	$\frac{1}{s i n 2 ϕ} = \frac{1}{2 t a n ϕ . {c o s}^{2} \emptyset} = \frac{1 + {t a n}^{2} ϕ}{2 t a n \emptyset} = \frac{1}{2} (\frac{1}{t a n \emptyset} + t a n ϕ)$ $\Rightarrow t a n ϕ = \frac{2 h}{a} \Rightarrow \frac{1}{s i n 2 ϕ} = \frac{h}{a} + \frac{a}{4 h}! ?$
	Commented by mr W last updated on 25/Sep/23

	$y e s, y o u a r e r i g h t s i r!$
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