Step 1: List the given shovel data.
We are given bucket capacity $= 20\ \text{m}^3$, fill factor $f = 0.8$, swell factor $s = 0.5$, and cycle time $t = 60$ seconds, and we want the hourly output volume.
Step 2: Find how much material each bucket load actually carries.
The fill factor tells us the bucket rarely loads to its full rated capacity, so the loose material picked up per dig is $20 \times 0.8 = 16\ \text{m}^3$ of loose (swelled) material.
Step 3: Bring in the swell factor to convert to the volume basis asked for.
Applying the swell factor of 0.5 to this loose volume gives $16 \times 0.5 = 8\ \text{m}^3$ per bucket load on the basis the problem intends.
Step 4: Scale up to a full hour and conclude.
In one hour of 3600 seconds, the number of cycles is $\dfrac{3600}{60} = 60$ cycles, so the hourly volume carried is $8 \times 60 = 480\ \text{m}^3$ per hour.
\[ \boxed{480\ \text{m}^3} \]