As given that, speed of packets $(u)=125 \mathrm{~m} / \mathrm{s}$
Height of the hill, $h=500 \mathrm{~m}$
To cross the hill by packet the vertical component of the (Speed of packet) should be sufficient to cross such height of $500 \mathrm{~m}$ and distance betwen hill and canon must be half the range of packet.
$$
\begin{aligned}
&v^2=u^2+2 g h \\
&0=u_x^2-2 g h \\
&u_y=\sqrt{2 g h}=\sqrt{2 \times 10 \times 500} \\
&u_y=100 \mathrm{~m} / \mathrm{s}
\end{aligned}
$$
But, $u^2=u_x^2+u_y^2$
So, Horizontal component of initial velocity,
$$
\begin{aligned}
&u_x=\sqrt{u^2-u_y^2}=\sqrt{(125)^2-(100)^2} \\
&u_x=75 \mathrm{~m} / \mathrm{sec}
\end{aligned}
$$
Time taken to reach the top of the hill,
$$
l=\sqrt{\frac{2 h}{g}}=\sqrt{\frac{2 \times 500}{10}}=10 \mathrm{sec}
$$
Time taken to reach the ground from the top of the hill $t^{\prime}=$ $t=10 \mathrm{~s}$. (Horizontal distance travelled in $10 \mathrm{sec}$ ).
$$
x=u_x \times t=75 \times 10=750 \mathrm{~m}
$$
So, distance through which canon has to be moved $=800$ $-750=50 \mathrm{~m}$
Speed with which canon can $m o v e=2 \mathrm{~m} / \mathrm{s}$
So, time taken by canon $=\frac{50}{2} \Rightarrow t^{\prime \prime}=25 \mathrm{~s}$
Hence, total time taken by a packet (from $800 \mathrm{~m}$ away from hill) to reach on the (otherside of) ground $=t^{\prime \prime}+\mathrm{t}+\mathrm{t}^{\prime}=25$ $+10+10=45 \mathrm{~s}$