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Q.

The trajectory of a projectile in a vertical plane is $y = α x - β x^{2},$ where $α$ and $β$ are constants and x & y are respectively the horizontal and vertical distances of the projectile from the point of projection. The angle of projection $θ$ and the maximum height attained H are respectively given by :

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a

$\tan^{- 1} α, \frac{4 α^{2}}{β}$

b

$\tan^{- 1} (\frac{β}{α}), \frac{α^{2}}{β}$

c

$\tan^{- 1} α, \frac{α^{2}}{4 β}$

d

$\tan^{- 1} β, \frac{α^{2}}{2 β}$

answer is C.

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Detailed Solution

$y = α x - β x^{2} C o m p a r e w i t h s \tan d a r d f o r m, y = (\tan θ) x - (\frac{g}{2 u^{2} \cos^{2} θ}) x^{2} W e g e t, \tan θ = α \Rightarrow θ = \tan^{- 1} [α] a n d \frac{α^{2}}{4 β} = \frac{\tan^{2} θ}{4 (\frac{g}{2 u^{2} \cos^{2} θ})} = H$

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