Given : r=72m, θ= 78°4’ , μs=0.8, g= 10m/s2 , tanθ= tan78°4’ =5
\(V_{min} =\sqrt{rg(\frac{tanθ-μ_s}{1+μ_stanθ})}\)
\(V_{min}= \sqrt{72×10(\frac{5-0.8}{1+(0.8((5)})}\)
vmin = 24.588 m/s = 88.52 km/h
88.52 km/h will be the lower limit or minimum speed on this track.
Since the track is heavily banked θ > 45° there is no upper limit or maximum speed on this track.