Answer:

The speed of packets is 125 m/s, the height of the hill is 500 m, and the distance between the cannon and the foot of the hill is 800 m, according to the problem.

Ideally, the vertical component of the velocity should be kept to a minimum so that it spans the height of the slope in the least amount of time.

${u}_{y}=\sqrt{2gh}\ge \sqrt{2\times 10\times 500}\ge 100m/s$u_{y}=\sqrt{2 g h} \geq \sqrt{2 \times 10 \times 500} \geq 100 m / s

But

So, Horizontal component of initial velocity,

The amount of time it took the packet to get to the top of the slope,

$t=\sqrt{\frac{2h}{g}}=\sqrt{\frac{2\times 500}{10}}=10s$Time taken to reach the ground from the top of the hill . Horizontal distance travelled in ,

$x={u}_{x}\times t=75\times 10=750m$The distance through which the cannon must be moved is equal to 800 – 750 = 50 metres. The maximum speed at which a canon may move is 2 metres per second.

Time taken by canon

Total time taken by a packet to reach on the ground