Question-154661

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 \dots (1)

y = d \sin ϕ = v_{i} \sin θ_{i} t - \frac{1}{2} {g t}^{2} \dots (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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