At what angle of banking, speed at turn is = v √gγ?

At what angle of banking, speed at turn is = v √gγ?

Explanation

The speed at a banked turn is given by:

v = √(rg tan(θ))

Given v = √(rg), we can equate:

√(rg tan(θ)) = √(rg)

This simplifies to:

tan(θ) = 1

θ = arctan(1) = 45°.