Consider drawn figure, let a fighter plane drops the bomb $(t \mathrm{sec})$ to hit a target $T$ when it be at position $P$.
If, $\angle Q P T=\theta$
Speed of the plane,
$$
\begin{aligned}
u &=720 \mathrm{~km} / \mathrm{h} \\
&=720 \times \frac{5}{18} \mathrm{~m} / \mathrm{s}=200 \mathrm{~m} / \mathrm{s}
\end{aligned}
$$
Altitude of the plane $(Q T), h=1.5 \mathrm{~km}=1500 \mathrm{~m}$
If bomb hits the target after time $t$, then horizontal distance travelled by the bomb.
$$
h(P Q)=u \times t=200 t
$$
When $(a=0)$,
$$
\begin{aligned}
&h=u t+g t \quad(\because u=0) \\
&g=9.8
\end{aligned}
$$
Vertical distance travelled by the bomb,
$$
h(Q T)=\frac{1}{2} g t^2 \Rightarrow 1500=\frac{1}{2} \times 9.8 t^2
$$
So, $t^2=\frac{1500}{4.9}$
$$
\Rightarrow t=\sqrt{\frac{1500}{4.9}}=\sqrt{\frac{1500}{5}}=\sqrt{300}=10 \sqrt{3} \mathrm{sec} \text {. }
$$
Using value of $t$ in Eq. (i),
$$
P Q=200 \times 10 \sqrt{3}=2000 \sqrt{3} \mathrm{~m}
$$
So, $\tan \theta=\frac{Q T}{Q P}=\frac{1500}{(2000 \times \sqrt{3})}=0.433$
$$
=\tan 23^{\circ} 42
$$
$$
\therefore \theta=23^{\circ} 42^{\prime}
$$