If, for the sake of argument, it turns out to be a nickle-iron asteroid, what sort of crater (on land) might it produce?
I'm not sure of the asteroid's speed, but assuming a typical 17km/s and other typical specifics (such as impact angle), your hypothetical gets this:
Distance from Impact: 1.00 km = 0.62 miles
Projectile Diameter: 11.00 m = 36.08 ft = 0.01 miles
Projectile Density: 8000 kg/m3
Impact Velocity: 17.00 km/s = 10.56 miles/s
Impact Angle: 45 degrees
Target Density: 2500 kg/m3
Target Type: Sedimentary Rock
Energy before atmospheric entry: 8.06 x 1014 Joules = 192.46 KiloTons TNT
The average interval between impacts of this size somewhere on Earth is 30.9 years
The projectile begins to breakup at an altitude of 13700 meters = 44800 ft
The projectile bursts into a cloud of fragments at an altitude of 7180 meters = 23500 ft
The residual velocity of the projectile fragments after the burst is 7.28 km/s = 4.52 miles/s
The energy of the airburst is 6.58 x 1014 Joules = 0.16 x 10^0 MegaTons.
Large fragments strike the surface and may create a crater strewn field. A more careful treatment of atmospheric entry is required to accurately estimate the size-frequency distribution of meteoroid fragments and predict the number and size of craters formed.
So the result is a 160 Kiloton airburst at just over 7km. Airblast is not noticeable (The overpressure is less than 1 Pa)
Using porous rock instead of iron, it bursts at 36.4km with 140 Kilotons of energy. Neither scenario is dangerous, and if I remember correctly, last year, an asteroid of similar size exploded over Indonesia. It would be more of a scientific interest than a danger, even if iron.
The initial estimate included the possibility that it was 25m. An iron asteroid of this size is tended to where it can strike the ground and create a significant crater.