I have seen a relationship in the curve path of a thrown object at β while the total passed distance D_v and highest point had passedD_u then β = arctan(((4D_u )/D_v )) but cant find the proof. I would like to say would anyone like to proove it?then please.


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Question Number 217148 by Marzuk last updated on 03/Mar/25

$I h a v e s e e n a r e l a t i o n s h i p i n t h e c u r v e$ $p a t h o f a t h r o w n o b j e c t a t β$ $w h i l e t h e t o t a l p a s s e d d i s t a n c e D_{v} a n d$ $h i g h e s t p o i n t h a d {p a s s e d D}_{u}$ $t h e n β = a r c t a n (\frac{4 D_{u}}{D_{v}})$ $b u t c a n t f i n d t h e p r o o f .$ $I w o u l d l i k e t o s a y w o u l d a n y o n e l i k e$ $t o p r o o v e i t ? t h e n p l e a s e .$
	Answered by mr W last updated on 03/Mar/25

	Commented by mr W last updated on 03/Mar/25

	$t h e t r a c k o f a t h r o w n o b j e c t i s a$ $p a r a b o l a (r e d c u r c e i n d i a g r a m) .$ $f o r a p a r a b o l a t h e r e a r e t h e$ $g e o m e t r i c a l r e l a t i o n s h i p s a s s h o w n .$ $\tan β = \frac{2 D_{v}}{0.5 D_{u}} = \frac{4 D_{v}}{D_{u}}$ $\Rightarrow β = a r c t a n (\frac{4 D_{v}}{D_{u}})$ $c e r t a i n l y w e c a n a l s o p r o v e t h i s$ $u s i n g t h e e q u a t i o n s f o r p r o j e c t i l e$ $m o t i o n, s e e b e l o w .$
	Answered by mr W last updated on 03/Mar/25

	Commented by Marzuk last updated on 03/Mar/25

	$M u c h a p p r e c i a t e d! I t m i g h t n e e d s o m e$ $m o r e e x p a n s i o n t o p u b l i s h i t . I w i l l$ $d o i t .$ $A t t h e c o n c l u s i o n, t h a n k s f o r t h e k e y$ $p r o o f .$
	Commented by mr W last updated on 03/Mar/25

	$x = (u \cos β) t$ $y = (u \sin β) t - \frac{{g t}^{2}}{2}$ $a t y = y_{m a x} = D_{v} :$ $\frac{d y}{d t} = u \sin β - g t = 0 \Rightarrow t = \frac{u \sin β}{g} = t_{1}$ $y ∣_{t = t_{1}} = D_{v} = (u \sin β) \times (\frac{u \sin β}{g}) - \frac{g}{2} {(\frac{u \sin β}{g})}^{2}$ $\Rightarrow D_{v} = \frac{u^{2} \sin^{2} β}{2 g} . . . (i)$ $a t y = 0, i . e . p o i n t B :$ $(u \sin θ) t - \frac{{g t}^{2}}{2} = 0 \Rightarrow t = t_{2} = \frac{2 u \sin β}{g} = 2 t_{1}$ $x ∣_{t = t_{2}} = D_{u} = (u \cos β) \times (\frac{2 u \sin β}{g})$ $\Rightarrow D_{u} = \frac{2 u^{2} \sin β \cos β}{g} . . . (i i)$ $(i) / (i i) :$ $\frac{D_{v}}{D_{u}} = \frac{\sin β}{4 \cos β} = \frac{\tan β}{4}$ $\Rightarrow β = a r c t a n (\frac{4 D_{v}}{D_{u}})$
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