Suppose that the fighter plane is flying horizontally with a speed v at the height OA = 1.5 km. The point O represents the position of the anti-aircraft gun.
Let u be the velocity of the shell andq, its inclination with the vertical. The shell hits the fighter plane at the point B as shown in Fig.
Suppose that the shell hits the plane after a time t. Then, the horizontal distance travelled by the fighter plane in time t with velocity v isequal to the horizontal distance covered by the shell in time t with ux, the x-component of its velocity i.e.
vt=ux×t or vt=u‌sin‌θ×t or sin‌θ=

v

u

Here, v=720kmh−1=200ms−1 and u=600ms−1
The minimum altitude at which the pilot should fly to avoid being hit,
H=

u2sin2(90−θ)

2g

=

u2cos2θ

2g

=

(600)2×(cos‌19.5°)2

2×10

=16000m=16km [∵sin‌θ=

1

3

(cos‌θ)=

58

3

]