Height of the hill = 500 m

To cross the hill

$\begin{array}{l}{\mathrm{u}}_{\mathrm{y}}\ge \sqrt{2\mathrm{gh}}\\ \ge \sqrt{2\times 10\times 500}\\ \ge 100\mathrm{m}/\mathrm{s}\\ \mathrm{But} {\mathrm{u}}^2={\mathrm{u}}_{\mathrm{x}}^2+{\mathrm{u}}_{\mathrm{y}}^2\end{array}$

Step 2: Calculate the horizontal component of velocity.

Given, speed of packets = 125 m/s
$\therefore$ The horizontal component of the initial velocity.
$u_x\mathit{ }= \sqrt{u^2-u_{y\mathit{ }}^2} = \sqrt{(125{)}^2-(100{)}^2} = 75 m/s$

Step 3: Calculate time is taken to reach the top of the hill.

$t\mathit{ }= \sqrt{\frac{2h}{g}} = \sqrt{\frac{2\times 500}{10}} = 10s$

Step 4: Calculate the horizontal distance traveled by the canon.

Time is taken to reach the ground from the top of the hill t'= t = 10s. Horizontal distance traveled in 10 s
$x\mathit{ }= u_x\times t\mathit{ }= 75\times 10 = 750 m$

$\therefore$Distance through which canon has to be moved = 800 - 750 = 50 m

Step 5: Calculate the total time taken by the packet to reach the ground.
The speed with which canon can move = 2 m/s
$\therefore \text{ Time taken by canon }=\frac{50}{2}\Rightarrow t\mathit{"}=25s$
$\therefore$Total time taken by a packet to reach on the ground =t" + t  +t'= 25 + 10 + 10 = 45 s