Question

(637) A cyclist is riding on a horizontal frozen meadow. He is turning so that he is tilted at an angle of 60°. Radius of curvature of the bend is 100m. What is the maximum speed of the cyclist?

Thank you for your time!
Since you do not specify, I will assume 60 degrees from horizontal
tan 60 = mg/(m v^2/r)

v^2/r = g/tan 60

v^2 = g r/tan 60

v = sqrt(981/tan 60)

v = 23.8 m/s

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