volcano exclusion zone

A couple of months ago a news item on BBC World News was that a volcano in Indonesia was sending ash and rocks 4 kms up into the atmosphere, and the authorities had evacuated the surrounding area and declared an exclusion zone with a 8 kms radius.

The ratio of 2 didn't appear to be a coincidence so I checked my physics book.
Sure enough, if the maximum height (for a projectile sent straight up) is v^2/(2g), then the maximum range is v^2/g, the latter corresponding to a 45 degree inclination, v is the initial velocity and g the acceleration due to gravity.

However, this neat picture based on my memory is contradicted by:

http://www.volcanolive.com/volcanolive.html

a) Residents living in 10 villages within an 8 km radius of Gamkonora volcano in Indonesia have been advised to evacuate after an increase in activity. Frequent tremors have been recorded at Gamhonora's crater since Sunday. Ash was emitted to an altitude of 8000 ft today. The volcano alert status has been raised to maximum level 4 (7th July 2007)

b) 12.77 N, 124.05 E, summit elevation 1565 m, stratovolcano
Tuesday 31st July 2007
Bulusan volcano erupted at 9:37 am (local time) today, covering nearby villages with ash. The eruption frightened residents who ran out of their houses. The ash column reached a height of 5km and drifted NW. The military is expected to enforce an evacuation of people from areas close to Bulusan. There is a 4 km permanent exclusion zone around the volcano.

So was my memory in error? Is the simple derivation relevant or irrelevant? Right now: I dunno.