The angle of banking is given by,
tan θ = \(\frac {v^2}{rg}\)
Here, v = \(\frac{ 45000}{60 \times60}\)
= 12.5 ms-1 and
r = 52 m
∴ tanθ = \(\frac{ (12.5)^2}{52\times 9.8}\)
= 0.3066
θ = tan-1 (0.3066)
= 17°2’
If ‘h’ is the height of the outer rail above the inner one, then
h = 1.2'sin 17.05°
= 0.352 m.