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Question Number 195870 by mr W last updated on 12/Aug/23

Commented by mr W last updated on 12/Aug/23

$f i n d t h e m i n i m u m s p e e d w i t h w h i c h$ $t h e b a l l i s l a u n c e d f r o m p o i n t A s u c h$ $t h a t i t c a n r e t u r n t o t h i s p o i n t a g a i n$ $a f t e r r e b o u n d e d a t p o i n t B .$ $a s s u m e t h e c o e f f i c i e n t o f r e s t i t u t i o n$ $i s e .$
Commented by liuxinnan last updated on 12/Aug/23

$i t a s i f h a s n o t t h e m i n v_{0}$ $b u t i t h a s a a s s u r e d a n g l e$ $b e t w e e n t h e v_{0} a n d t h e i n c l i n e$
Commented by mahdipoor last updated on 13/Aug/23

$It seems that the question does not haven$ $minimum speed, instead, i find relationship$ $between tangential and vertical speed$ $(relative to the surface) :$ $\frac{V_{t}}{V_{v}} = \frac{v}{u} = \frac{e^{2} + 2 e + 1}{e + 1} t a n θ = (e + 1) t a n θ$
Commented by mr W last updated on 13/Aug/23

${i t}^{'} s c o r r e c t . t h e r e i s n o m i n i m u m f o r$ $s p e e d u, s i n c e t h e r e b o u n d p o i n t B$ $i s n o t f i x e d .$ $b u t i g o t \tan φ = \frac{1}{(e + 1) \tan θ}$
	Answered by mr W last updated on 13/Aug/23

	Commented by mr W last updated on 13/Aug/23

	$m o t i o n f r o m A t o B :$ $0 = u \sin φ t - \frac{g \cos θ}{2} t^{2}$ $\Rightarrow t = \frac{2 u \sin φ}{g \cos θ}$ $s = u \cos φ t - \frac{g \sin θ}{2} t^{2}$ $s = u \cos φ \times \frac{2 u \sin φ}{g \cos θ} - \frac{g \sin θ}{2} \times {(\frac{2 u \sin φ}{g \cos θ})}^{2}$ $s = \frac{u^{2}}{g \cos θ} (\sin 2 φ - 2 \sin^{2} φ \tan θ)$ $U_{x} = u \cos φ - g \sin θ \times \frac{2 u \sin φ}{g \cos θ}$ $= u (\cos φ - 2 \tan θ \sin φ)$ $U_{y} = u \sin φ - g \cos θ \times \frac{2 u \sin φ}{g \cos θ}$ $= - u \sin φ$ $V_{x} = U_{x} = u (\cos φ - 2 \tan θ \sin φ)$ $V_{y} = - {e U}_{y} = e u \sin φ$ $m o t i o n f r o m B t o A :$ $t = \frac{2 e u \sin φ}{g \cos θ}$ $- s = u (\cos φ - 2 \tan θ \sin φ) t - \frac{g \sin θ}{2} t^{2}$ $- s = u (\cos φ - 2 \tan θ \sin φ) \times \frac{2 e u \sin φ}{g \cos θ} - \frac{g \sin θ}{2} \times {(\frac{2 e u \sin φ}{g \cos θ})}^{2}$ $- \frac{u^{2}}{g \cos θ} [\sin 2 φ - 2 \sin^{2} φ \tan θ] = u (\cos φ - 2 \tan θ \sin φ) \times \frac{2 e u \sin φ}{g \cos θ} - \frac{g \sin θ}{2} \times {(\frac{2 e u \sin φ}{g \cos θ})}^{2}$ $- \sin 2 φ + 2 \sin^{2} φ \tan θ = 2 e \sin φ (\cos φ - 2 \tan θ \sin φ) - 2 e^{2} \tan θ \sin^{2} φ$ $(e + 1) \sin φ \cos φ = {(e + 1)}^{2} \tan θ \sin^{2} φ$ $\Rightarrow \tan φ = \frac{1}{(e + 1) \tan θ}$ $s = \frac{u^{2}}{g \cos θ} (\sin 2 φ - 2 \sin^{2} φ \tan θ)$ $\Rightarrow \frac{u^{2}}{g s} = [{(e + 1)}^{2} + \frac{1}{\tan^{2} θ}] \frac{\sin θ}{2 e}$
	Commented by mr W last updated on 13/Aug/23

	Commented by mr W last updated on 13/Aug/23

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