First slide

Projection Under uniform Acceleration

Question

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

Moderate

A

$\tan^{- 1} 3 ∣$
B

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

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

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

Solution

$H = \frac{u^{2} \sin^{2} θ}{2 g} or 80 = \frac{u^{2} \sin^{2} θ}{2 \times 10} u^{2} \sin^{2} θ = 1600 or usinθ = 40 {ms}^{- 1} Horizontal velocity = ucosθ = 3 \times 30 = 90 {ms}^{- 1} \frac{usinθ}{ucosθ} = \frac{40}{90} tanθ = \frac{4}{9} θ = \tan^{- 1} (\frac{4}{9})$

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