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 $u_x$, the x-component of its velocity i.e.$v t = u_x \times t$ or $v t = u \sin \theta \times t$ or $\sin \theta = \frac{v}{u}$Here, $v = 720\,\mathrm{km}\,\mathrm{h}^{-1} = 200\,\mathrm{m}\,\mathrm{s}^{-1}$ and $u = 600\,\mathrm{m}\,\mathrm{s}^{-1}$The minimum altitude at which the pilot should fly to avoid being hit,$H = \frac{u^{2} \sin^{2}(90-\theta)}{2g} = \frac{u^{2} \cos^{2} \theta}{2g}$$= \frac{(600)^{2} \times (\cos 19.5^{\circ})^{2}}{2 \times 10}$ $= 16000\,\mathrm{m} = 16\,\mathrm{km}$ $\left[\because \sin\theta = \frac{1}{3}(\cos\theta) = \frac{58}{3}\right]$