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Question Number 196697 by mr W last updated on 29/Aug/23

Commented by mr W last updated on 29/Aug/23
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Commented by mr W last updated on 29/Aug/23

$a b a l l i s r e l e a s e d f r o m a h e i g h t o f h$ $a b o v e a l o n g i n c l i n e d p l a n e a s s h o w n .$ $i f t h e c o e f f i c i e n t o f r e s t i t u t i o n i s e,$ $f i n d t h e v e l o c i t y o f t h e b a l l j u s t a f t e r$ $t h e n^{t h} c o l l i s i o n .$
Commented by mahdipoor last updated on 29/Aug/23
$obtained fron the same question as before: { ((v_i =v_1 e^(i−1) )),((u_i =u_1 −2v_1 (1+e+e^2 +...+e^(i−2) )tanθ)) :} in this case ⇒ { ((v_1 =e(√(agh))×cosθ)),((u_1 =−(√(2gh))×sinθ)) :} { ((v_i =(√(2gh)).cosθ×e^i )),((u_i =−(√(2gh)).sinθ(((1+3e−2e^i )/(1−e))))) :}$
$o b t a i n e d f r o n t h e s a m e q u e s t i o n a s b e f o r e :$ ${\begin{cases} v_{i} = v_{1} e^{i - 1} \\ u_{i} = u_{1} - 2 v_{1} (1 + e + e^{2} + . . . + e^{i - 2}) t a n θ \end{cases}$ $i n t h i s c a s e \Rightarrow {\begin{cases} v_{1} = e \sqrt{a g h} \times c o s θ \\ u_{1} = - \sqrt{2 g h} \times s i n θ \end{cases}$ ${\begin{cases} v_{i} = \sqrt{2 g h} . c o s θ \times e^{i} \\ u_{i} = - \sqrt{2 g h} . s i n θ (\frac{1 + 3 e - 2 e^{i}}{1 - e}) \end{cases}$
	Answered by mr W last updated on 30/Aug/23

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