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Question Number 53811 by ajfour last updated on 26/Jan/19

Commented by ajfour last updated on 26/Jan/19

$F i n d u_{m i n} a n d c o r r e s p o n d i n g θ f o r$ $t h i s t o b e p o s s i b l e .$ $(e i s t h e c o e f f i c i e n t o f r e s t i t u t i o n$ $f o r c o l l i s i o n o f b a l l w i t h g r o u n d$ $a n d w i t h w a l l a s w e l l .)$
	Answered by mr W last updated on 27/Jan/19

	$a t p o i n t o n g r o u n d :$ $v_{1, x, i} = v_{1, x, f} = u \cos θ$ $v_{1, y, i} = u \sin θ$ $v_{1, y, f} = e u \sin θ$ $a t p o i n t o n w a l l :$ $v_{2, x, i} = v_{1, x, f} = u \cos θ$ $t = \frac{b}{v_{2, x, f}} = \frac{b}{u \cos θ}$ $v_{2, y, i} = v_{1, y, f} - g t = e u \sin θ - \frac{g b}{u \cos θ}$ $h = v_{1, y, f} t - \frac{1}{2} {g t}^{2} = e u \sin θ \times \frac{b}{u \cos θ} - \frac{g}{2} {(\frac{b}{u \cos θ})}^{2}$ $\Rightarrow h = e b \tan θ - \frac{{g b}^{2}}{2 u^{2} \cos^{2} θ}$ $v_{2, x, f} = {e v}_{2, x, i} = e u \cos θ$ $v_{2, y, f} = v_{2, y, i} = e u \sin θ - \frac{g b}{u \cos θ}$ $a t p o i n t o n g r o u n d a g a i n :$ $t = \frac{b}{v_{2, x, f}} = \frac{b}{e u \cos θ}$ $- h = v_{2, y, f} t - \frac{1}{2} {g t}^{2} = (e u \sin θ - \frac{g b}{u \cos θ}) \frac{b}{e u \cos θ} - \frac{g}{2} {(\frac{b}{e u \cos θ})}^{2}$ $\Rightarrow h = - b \tan θ + \frac{{g b}^{2}}{u^{2} \cos^{2} θ} (\frac{1}{e} + \frac{1}{2 e^{2}})$ $\Rightarrow e b \tan θ - \frac{{g b}^{2}}{2 u^{2} \cos^{2} θ} = - b \tan θ + \frac{{g b}^{2}}{u^{2} \cos^{2} θ} (\frac{1}{e} + \frac{1}{2 e^{2}})$ $\Rightarrow (e + 1) \tan θ = \frac{g b}{u^{2} \cos^{2} θ} (\frac{1}{2} + \frac{1}{e} + \frac{1}{2 e^{2}})$ $\Rightarrow (e + 1) \tan θ = \frac{(1 + \tan^{2} θ) g b}{2 u^{2}} {(\frac{1 + e}{e})}^{2}$ $\Rightarrow u^{2} = \frac{g b}{2} (\frac{1 + e}{e^{2}}) (\tan θ + \frac{1}{\tan θ})$ $s i n c e \tan θ + \frac{1}{\tan θ} ⩾ 2 (‘ ‘ =^{″} i s a t θ = 45 °)$ $\Rightarrow u_{m i n}^{2} = g b (\frac{1 + e}{e^{2}})$ $\Rightarrow u_{m i n} = \frac{1}{e} \sqrt{(1 + e) g b}$ $a t θ = 45 °$ $============$ $a l t e r n a t i v e w a y :$ $a f t e r r e b o u n d f r o m t h e g r o u n d a n d$ $b a c k :$ $t = \frac{b}{u \cos θ} + \frac{b}{e u \cos θ} = \frac{(1 + e) b}{e u \cos θ}$ $0 = e u \sin θ t - \frac{1}{2} {g t}^{2}$ $\Rightarrow e u \sin θ = \frac{1}{2} g t = \frac{(1 + e) g b}{2 e u \cos θ}$ $\Rightarrow u^{2} = \frac{(1 + e) g b}{e^{2} \sin 2 θ}$ $\Rightarrow u = \frac{1}{e} \sqrt{\frac{(1 + e) g b}{\sin 2 θ}}$ $\Rightarrow u_{m i n} = \frac{\sqrt{(1 + e) g b}}{e} a t θ = 45 °$
	Commented by ajfour last updated on 26/Jan/19

	$t h a n k s S i r . B e a u t i f u l r e s u l t, {i s n}^{'} t ?$
	Commented by mr W last updated on 26/Jan/19

	${i t}^{'} s i n d e e d a v e r y n i c e q u e s t i o n!$
	Commented by mr W last updated on 26/Jan/19

	${t h a t}^{'} s i t! t h a n k y o u f o r c h e c k i n g s i r!$
	Commented by mr W last updated on 26/Jan/19

	$a s c h i l d r e n w e o f t e n t r i e d t o p l a y$ $t h i s w i t h a t e n n i s b a l l t o s e e i f w e c o u l d$ $g e t t h e b a l l b a c k t o t h e s a m e p o i n t o n t h e$ $g r o u n d .$
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