A fighter plane flying horizontally at an altitude of $2 k m$ with a speed of $540 k m / h$ passes directly overhead an anti-aircraft gun. If the gun can fire a bullet at the muzzle speed of $500 m s^{- 1}$ , at what angle with the vertical should the gunfire so that the bullet hits the plane?

{cos}^{- 1} (\frac{3}{10})

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{sin}^{- 1} (\frac{3}{10})

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{tan}^{- 1} (\frac{3}{10})

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{tan}^{- 1} (1)

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Solution

The correct option is A ${sin}^{- 1} (\frac{3}{10})$
for bullet the muzzle velocity $= 500$ m/s $= 500 \times \frac{10^{- 3} k m}{\frac{1}{60 \times 60} h r} = 1800$ km/hr
Assume bullet fired at a angle $θ$ and velocity of gun is zero
$u_{h} = 1800 sin θ$ km/hr
$u_{v} = 1800 cos θ$ km/hr
where $u_{h}$ and $u_{v}$ are horizontal and vertical velocities
Given: $S_{v} = 2 k m$
Let time taken be $t$ hr
Dist. traveled by plane, $S = 540 \times t$ km
This is equal to range of bullet
$∴ R = u_{h} \times t = 540 \times t$
$\Rightarrow 1800 sin θ \times t = 540 \times t$
$\Rightarrow sin θ = \frac{3}{10}$
$θ = {sin}^{- 1} (\frac{3}{10})$

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A fighter plane flying horizontally at an altitude of 2km with a speed of 540km/h passes directly overhead an anti-aircraft gun. If the gun can fire a bullet at the muzzle speed of 500ms−1, at what angle with the vertical should the gunfire so that the bullet hits the plane?

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