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Question Number 186416 by ajfour last updated on 04/Feb/23

Commented by ajfour last updated on 04/Feb/23

$Q . 186345$
	Answered by ajfour last updated on 04/Feb/23
	$x=pt y=h+((gd^2 )/(2p^2 ))=a ..(i) e(√(2gh))=(√(2gb)) ⇒ h=(b/e^2 ) ..(ii) (√((2h)/g))−(√((2(h−a))/g))=(x/p) ...(iii) (x/p)+(√((2b)/g))=(c/p) ....(iv) q=usin θ=g(√((2(h−a))/g)) ...(v) (iii)+(iv) gives (√h)−(√(h−a))+(√b)=(c/p)(√(g/2)) but h=(b/e^2 ) , hence p=((√((gc^2 )/(2b)))/((1+(1/e)−(√((1/e^2 )−(a/b)))))) q=usin θ=(√(2gb((1/e^2 )−(a/b)))) from eq.(iv) x=c−p(√((2b)/g)) x=c{1−(1/((1+(1/e)−(√((1/e^2 )−(a/b))))))} u=(√(2gb))(√(((1/e^2 )−(a/b))+(((((2c)/b))^2 )/((1+(1/e)−(√((1/e^2 )−(a/b))))))))$
	$x = p t$ $y = h + \frac{{g d}^{2}}{2 p^{2}} = a . . (i)$ $e \sqrt{2 g h} = \sqrt{2 g b}$ $\Rightarrow h = \frac{b}{e^{2}} . . (i i)$ $\sqrt{\frac{2 h}{g}} - \sqrt{\frac{2 (h - a)}{g}} = \frac{x}{p} . . . (i i i)$ $\frac{x}{p} + \sqrt{\frac{2 b}{g}} = \frac{c}{p} . . . . (i v)$ $q = u \sin θ = g \sqrt{\frac{2 (h - a)}{g}} . . . (v)$ $(i i i) + (i v) g i v e s$ $\sqrt{h} - \sqrt{h - a} + \sqrt{b} = \frac{c}{p} \sqrt{\frac{g}{2}}$ $b u t h = \frac{b}{e^{2}}, h e n c e$ $p = \frac{\sqrt{\frac{{g c}^{2}}{2 b}}}{(1 + \frac{1}{e} - \sqrt{\frac{1}{e^{2}} - \frac{a}{b}})}$ $q = u \sin θ = \sqrt{2 g b (\frac{1}{e^{2}} - \frac{a}{b})}$ $f r o m e q . (i v)$ $x = c - p \sqrt{\frac{2 b}{g}}$ $x = c {1 - \frac{1}{(1 + \frac{1}{e} - \sqrt{\frac{1}{e^{2}} - \frac{a}{b}})}}$ $u = \sqrt{2 g b} \sqrt{(\frac{1}{e^{2}} - \frac{a}{b}) + \frac{{(\frac{2 c}{b})}^{2}}{(1 + \frac{1}{e} - \sqrt{\frac{1}{e^{2}} - \frac{a}{b}})}}$
	Commented by mr W last updated on 04/Feb/23

	$g r e a t s o l u t i o n s i r!$
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