[Bob]

This could be a freshman physics question. To drop 70 feet takes sqrt( 70 ft/ (2 * 32 ft/sec2) ) = 1.05 sec

I'm assuming that you're calculating from the 85 foot top of the embankment to the 15 foot high impact with the tree, right?

IANAP (I am not a physicist), but I think there's a complication in that going up the embankment would have launched the car skyward from the 85 foot level. Depending on the steepness the embankment, the car could have gone significantly higher than the top of the embankment.

[dr_lew]

By the Google Maps image, the north end of the runway sits on top of a mound which drops off on three sides to form the embankment.

Also, I erred by a factor of 2 ( Arrgh. ) s = 1/2 g t^2, so t=sqrt(2s/g) not sqrt(s/2g). That means it had about 2 seconds to go 200 feet, requiring a speed of only 65 mph, by this estimate. This is more consistent with braking and sliding after a high speed run.

[Atlas Sneezed]

I think there’s a complication in that going up the embankment would have launched the car skyward from the 85 foot level.

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I suspect that the embankment slopes downward from the runway level, not upward (which would be an insane obstacle to put at the end of a runway to be hit by planes attempting take-off!)