$A$ fighter plane is flying horizontally at an altitude of $1.5\, km$ with a speed of $720\, km/h$. At what angle of sight (with respect to the horizontal) should the pilot drop the bomb when the target is seen,in order to hit the target?

(N/A) Let the fighter plane be at point $P$ when it drops a bomb to hit a target $T$. The target is vertically below point $P'$ on the flight path.
Speed of the plane $u = 720\, km/h = 720 \times \frac{5}{18}\, m/s = 200\, m/s$.
Height of the plane $h = P'T = 1.5\, km = 1500\, m$.
Let $t$ be the time taken by the bomb to hit the target. The vertical distance covered is $h = \frac{1}{2}gt^2$.
$1500 = \frac{1}{2} \times 9.8 \times t^2 \implies t^2 = \frac{3000}{9.8} \approx 306.12$.
$t = \sqrt{306.12} \approx 17.5\, s$.
The horizontal distance covered by the bomb in this time is $x = P'T = u \times t = 200 \times 17.5 = 3500\, m$.
The angle of sight $\theta$ with the horizontal is given by $\tan \theta = \frac{\text{Vertical distance}}{\text{Horizontal distance}} = \frac{P'T}{PP'} = \frac{1500}{3500} = \frac{3}{7} \approx 0.4286$.
$\theta = \tan^{-1}(0.4286) \approx 23.2^{\circ}$.