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Q.

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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a

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

b

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

c

$45^{\circ}$

d

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

answer is B.

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Detailed Solution

Question Image

$x = Vt = usin θ t$

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

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

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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 ?