A fighter plane flying horizontal at an altitude of 1.5 Km with speed of 720 Km $h^{- 1}$ passes directly overhead an anticarft gun. At what angle from the vertical should the gun be fired from the shell with muzzle speed 400 $m s^{- 1}$ to hit the plane in shortest time.?

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14.47

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Solution

The correct option is B 14.47
consider the aircraft gun is launched at an angle $θ$ from vertical at a speed of $u = 400 m / s$ . it follows a projectile motion
consider they collide at a time t.
in time $t$ ,fighter plane travels.
$d = v t$
also for aircraft gun
$u sin θ t = d$
that gives
$v t = u sin t h e t a t$
$\Rightarrow u sin θ = v \Rightarrow sin θ = \frac{v}{u} = \frac{720 \times \frac{5}{18}}{400} = \frac{4 \times 5}{40} = \frac{1}{4}$
$\Rightarrow θ = {sin}^{- 1} (\frac{1}{4}) ≃ 14.470$
so at $θ = 14.470$ ,its unit the plane in minimum time

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A fighter plane flying horizontal at an altitude of 1.5 Km with speed of 720 Km h−1 passes directly overhead an anticarft gun. At what angle from the vertical should the gun be fired from the shell with muzzle speed 400 ms−1 to hit the plane in shortest time.?

A fighter plane flying horizontal at an altitude of 1.5 Km with speed of 720 Km $h^{- 1}$ passes directly overhead an anticarft gun. At what angle from the vertical should the gun be fired from the shell with muzzle speed 400 $m s^{- 1}$ to hit the plane in shortest time.?