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

Motion in a plane

Speed of the fighter plane = 720 km/h = 720 x (5/18) = 200 m/s

The altitude of the plane = 1.5 km

Velocity of the shell = 600 m/s

Sin θ = 200/600 = 1/3

θ = sin-1 (1/3) = 19.470

Let H be the minimum altitude

Using equation,

H = u2 sin2 (90 – θ)/2g

= (6002 cos2θ)/2g

= 6002 cos 2θ/(2 x 9.8)= {360000[(1+cos 2θ)/2]}/2g

= 360000[1+cos2 (19.470)/2]/2g

= 360000[ (1 + cos 38.94)/2]/(2 x 9.8)

= 360000 [ (1 + 0.778)/2]/19.6

= 360000 [(1.778/2)]/19.6

= (360000 x 0.889) /19.6

= 320040/19.6

= 16328 m

= 16.328 km

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