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A projectile has initially the same horizontal velocity as it would acquire if it had moved from rest with uniform acceleration of $3 m s^{- 2}$ for $0.5 \min$ . If the maximum height reached by it is $80 m$ , then the angle of projection is $(g = 10 m s^{- 2})$

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$\tan^{- 1} 3$

$\tan^{- 1} (3 / 2)$

$\tan^{- 1} (4 / 9)$

$\sin^{- 1} (4 / 9)$

answer is C.

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Detailed Solution

$H = \frac{u^{2} \sin^{2} θ}{2 g} or 80 = \frac{u^{2} \sin^{2} θ}{2 \times 10}$

$or u^{2} \sin^{2} θ = 1600 or u \sin θ = 40 {ms}^{- 1}$

$Horizontal velocity = u \cos θ = 3 \times 30 = 90 {ms}^{- 1}$

$\frac{u \sin θ}{u \cos θ} = \frac{40}{90}$

$⇒ \tan θ = \frac{4}{9}$

$⇒ θ = \tan^{- 1} (\frac{4}{9})$

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