A fighter plane flying horizontally at an altitude of 2 km with speed of 540 kmph passes directly over head an antiaircraft gun. If the gun can fire a bullet at the muzzle speed of $500 {ms}^{- 1}$ , at what angle with the vertical the gun should fire the bullet so that the bullet hits the plane ?

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${Sin}^{- 1} (3 / 10)$

$\tan^{- 1} (10 / 3)$

$45^{\circ}$

${Cos}^{- 1} (3 / 10)$

answer is B.

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Detailed Solution

$x = Vt = usin θ t$

$540 \times \frac{5}{10} = 500 \sin θ$

$\sin θ = \frac{3}{10} \Rightarrow θ = \sin^{- 1} (3 / 10)$

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