A particle is projected from point A. A fly is sitting at a tower at point B as shown in the figure. When particle passes from B, fly starts flying in horizontal direction with constant speed of 2 m/s. Both will meet at the same horizontal level at point C, then (take, g = 10 m/s2)

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A particle is projected from point A. A fly is sitting at a tower at point B as shown in the figure. When particle passes from B, fly starts flying in horizontal direction with constant speed of 2 m/s. Both will meet at the same horizontal level at point C, then (take, g = 10 m/s²)

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None of these

angle of projection is 45°

angle of projection is $\tan^{- 1} (\frac{9}{2})$

angle of projection is tan^-1 (2)

answer is A.

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Detailed Solution

$u \cos θ t = 2 t u \cos θ = 2 - - - (1) \begin{array}{l} y = xtan θ - \frac{{gx}^{2}}{2 u^{2} \cos^{2} θ} \\ \Rightarrow 4 = 2 \tan θ - \frac{5 (2)^{2}}{(2)^{2}} \\ \Rightarrow \tan θ = \frac{9}{2} \\ \Rightarrow θ = \tan^{- 1} (\frac{9}{2}) \end{array}$