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Question Number 91196 by ajfour last updated on 28/Apr/20

Commented by ajfour last updated on 28/Apr/20

$T h e b a l l h a s t o b e d r o p p e d o n$ $i n c i l i n e s o t h a t i t l a n d s i n t o$ $t h e b o x a n d y_{P} b e m i n i m u m .$ $F i n d x_{P} i n t e r m s o f b .$
	Answered by mr W last updated on 28/Apr/20

	Commented by mr W last updated on 28/Apr/20

	$l e t x_{P} = - a$ $u = \sqrt{2 g (y_{P} - a)}$ $a = \frac{1}{2} g {(\frac{a + b}{u})}^{2} = \frac{{(a + b)}^{2}}{4 (y_{P} - a)}$ $y_{P} = a + \frac{{(a + b)}^{2}}{4 a}$ $\frac{{d y}_{P}}{d a} = 1 + \frac{(a + b)}{2 a} - \frac{{(a + b)}^{2}}{4 a^{2}} = 0$ $5 a^{2} - b^{2} = 0$ $\Rightarrow a = \frac{b}{\sqrt{5}}$ $i . e . x_{P} = - \frac{b}{\sqrt{5}}$
	Commented by ajfour last updated on 28/Apr/20
	$i think you got a=b(2+(√5)) , Sir {(((a+b)/(2a)))+(1/2)}^2 =(((√5)/2))^2$
	$i t h i n k y o u g o t$ $a = b (2 + \sqrt{5}), S i r$ ${(\frac{a + b}{2 a}) + \frac{1}{2}}^{2} = {(\frac{\sqrt{5}}{2})}^{2}$
	Commented by mr W last updated on 28/Apr/20
	$no. i think a=(b/(√5)) is correct sir. {(((a+b)/(2a)))−(1/2)}^2 =(((√5)/2))^2$
	$n o . i t h i n k a = \frac{b}{\sqrt{5}} i s c o r r e c t s i r .$ ${(\frac{a + b}{2 a}) - \frac{1}{2}}^{2} = {(\frac{\sqrt{5}}{2})}^{2}$
	Commented by ajfour last updated on 28/Apr/20

	$y e s s i r, r e a l l y, a n d y o u$ $h a v e s o l v e d i t v e r y w i s e l y, S i r!$
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