I once fiddled around with a spreadsheet to investigate this scenario. I concluded from a straight-out departure, you need several thousand feet of runway beyond the rotation point. Otherwise, after the height loss in the turn, any expected glide ratio is insufficient to carry you back to the point of rotation.
Assume
Best climb of 60 kts = 1 NM / min = 6000 ft per min.
Climb rate: 600 fpm
Height lost in 180 deg turn: 300ft
(According to current wealth of experimental data: Thanks chaps!)
Glide ratio 10:1
1 Minute after Rotation
Dist = 6,000ft, Height = 600ft, Hgt after turn 300ft => gliding distance 3,000ft
2 Minutes after Rotation
Dist = 12,000ft, Height = 1,200ft, Hgt after turn 900ft => gliding distance 9,000ft
In each case the gliding distance is insufficient to backtrack distance covered – need another 3,000ft of runway beyond the point of rotation.
Neither light tail winds (10Kts), nor the trigonometry of 5 deg climb slopes materially affect the outcome.
...all rather academic, but stresses how difficult it is to make it back safely!