A fighter plane flying horizontally at an altitude of 1.5 km with a speed of 720 km/h passes directly overhead an anti-aircraft gun. At what angle from the vertical should the gun be fired for the shell with muzzle speed 600 m s-1 to hit the plane? At what minimum altitude should the pilot fly the plane to avoid being hit? (Take g = 10 m s-2 ).

Asked by Abhisek | 1 year ago |  194

##### Solution :-

Speed of the fighter plane = 720 km/h = 720 x ($$\dfrac{5}{18}$$) = 200 m/s

The altitude of the plane = 1.5 km

Velocity of the shell = 600 m/s

Sin θ = $$\dfrac{200}{600}$$ =$$\dfrac{1}{3}$$

θ = sin-1 ($$\dfrac{1}{3}$$) = 19.47°

Let H be the minimum altitude

Using equation,

H = u2 sin2 $$\dfrac{ (90 – θ)}{2g}$$

$$\dfrac{(600^2 cos^2θ)}{2g}$$

=$$\dfrac{(600^2 cos^2θ)}{2\times 9.8}$$$$\dfrac{360000[(1+cos 2θ)}{2}.\dfrac{1}{2g}$$

$$\dfrac{ 360000[1+cos2 (19.470)}{2}.\dfrac{1}{2g}$$

$$\dfrac{360000[ (1 + cos \;38.94)}{2}.\dfrac{1}{(2 \times 9.8)}$$

$$\dfrac{360000 [ (1 + 0.778)}{2}.\dfrac{1}{19.6}$$

$$\dfrac{ (360000 \times 0.889)}{19.6}$$

$$\dfrac{ 320040}{19.6}$$

= 16328 m = 16.328 km

Answered by Pragya Singh | 1 year ago

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