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Question Number 154661 by physicstutes last updated on 20/Sep/21

	Answered by mr W last updated on 20/Sep/21

	$x = d \cos ϕ = v_{i} \cos θ_{i} t . . . (1)$ $y = d \sin ϕ = v_{i} \sin θ_{i} t - \frac{1}{2} {g t}^{2} . . . (2)$ $f r o m (1) :$ $t = \frac{d \cos ϕ}{v_{i} \cos θ_{i}}$ $p u t t h i s i n t o (2) :$ $d \sin ϕ = v_{i} \sin θ_{i} \times \frac{d \cos ϕ}{v_{i} \cos θ_{i}} - \frac{1}{2} g {(\frac{d \cos ϕ}{v_{i} \cos θ_{i}})}^{2}$ $\sin ϕ = \sin θ_{i} \times \frac{\cos ϕ}{\cos θ_{i}} - \frac{g d}{2 v_{i}^{2}} \times \frac{\cos^{2} ϕ}{\cos^{2} θ_{i}}$ $\frac{g d}{2 v_{i}^{2}} \times \frac{\cos^{2} ϕ}{\cos θ_{i}} = \sin θ_{i} \cos ϕ - \cos θ_{i} \sin ϕ$ $\frac{g d}{2 v_{i}^{2}} \times \frac{\cos^{2} ϕ}{\cos θ_{i}} = \sin (θ_{i} - ϕ)$ $d = \frac{2 v_{i}^{2} \cos θ_{i} \sin (θ_{i} - ϕ)}{g \cos^{2} ϕ}$ $\frac{d}{d θ_{i}} (\frac{g d \cos^{2} ϕ}{2 v_{i}^{2}}) = \frac{d}{d θ_{i}} [\cos θ_{i} \sin (θ_{i} - ϕ)] = 0$ $- \sin θ_{i} \sin (θ_{i} - ϕ) + \cos θ_{i} \cos (θ_{i} - ϕ) = 0$ $\cos (2 θ_{i} - ϕ) = 0$ $\Rightarrow 2 θ_{i} - ϕ = \frac{π}{2}$ $\Rightarrow θ_{i} = \frac{π}{4} + \frac{ϕ}{2}$ $d_{m a x} = \frac{2 v_{i}^{2} \cos (\frac{π}{4} + \frac{ϕ}{2}) \sin (\frac{π}{4} - \frac{ϕ}{2})}{g \cos^{2} ϕ}$ $d_{m a x} = \frac{v_{i}^{2} {(\cos \frac{ϕ}{2} - \sin \frac{ϕ}{2})}^{2}}{g \cos^{2} ϕ}$ $d_{m a x} = \frac{v_{i}^{2} (1 - \sin ϕ)}{g \cos^{2} ϕ}$ $d_{m a x} = \frac{v_{i}^{2} (1 - \sin ϕ)}{g (1 - \sin^{2} ϕ)}$ $\Rightarrow d_{m a x} = \frac{v_{i}^{2}}{g (1 + \sin ϕ)}$
	Commented by mathdanisur last updated on 21/Sep/21

	$Very nice S er thankyou$
	Commented by physicstutes last updated on 21/Sep/21

	$+ brilliant +$
	Commented by Tawa11 last updated on 21/Sep/21

	$Weldone sir$
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