If you had the slightest clue about reality, you would understand the analogy is flawed.
A car did not hit the pole at 30 MPH.
Show me the light pole design considerations for aircraft impact in this particular case?
See ya in skewl.
Something much heavier than a car hit the lightpole at a far greater speed, and you want calculations to show that it will still break?
It also hit higher on the pole, crating a moment force as well as the shear forces a car would exert.
Very well.
Plane mass: 100 tonnes = 100,000 kg
Plane speed (conservative): 500 mph
Plane-pole impact height (approximate) = 45 feet
Pole height (representative): 55 feet
source
Pole mass (representative): 450 kg
source
Pole coupling ultimate tensile strength (representative): 221.5 kN
source
Pole coupling yielding strength (representative): 192 kN
source
Pole coupling ulimate restrained shear strength (representative): 24.5 kN
source
Pole coupling diameter (representative): 1 inche
source
Conversions:
Plane-pole impact

= 13.72 m
Pole height (h) = 16.76 m
Plane speed (v
p = 223.52 m/s
Coupling diameter (d) = 0.025 m
Since the plane impact high on the pole, the predominant force will be a moment.
For the pole to break, this moment force must cause more than 221.5 kN of force on the shear pins.
Note that there are typically 4 shear pins per pole. At most, the plane only needs to break 2 at a time.
These pins therefore provide a resistance force of 443 kN, at the base of the pole.
To break the pole, the plane must provide a certain minimum force at its impact point. Now things get fun.
I will assume the dimension ebtween the centre plane of the lightpole and the plane of the couplings is 0.10 m. Feel free to measure a local lightpole and re-do the work, if you wish.
This means the couplings provide a moment of 44.3 kN*m to resist breaking.
At a height of 13.72 m, the airplane need exert only 44.3/13.72 = 73 N of force to break the any two pins.
The plane masses 100,000 kg.
The acceleration caused by this exertion on such a mass is a = F/m = 73/100,000 = 0.00073 m/s
2.
Even if we double the moment by asusming all four pins must break simultaneosly, it still only requires a deceleration of 0.00146 m/s. Hardly noticible.
And at a speed of 223.5 m/s, the plane will only be impacting the lightpoles for a tiny fraction of a second.
The higher speed and higher mass of the plane result in far more force than a car an generate.
And cars have no problem taking out light standards.
I saw a minivan take one out, travelling at about 50 kph. It not only broke the couplings (as designed), but sent the pole flying 30 feet into the air.