Solution:
The angle of banking is given by \(sin\theta \approx \tan\theta = \frac{h}{w} = \frac{2}{20} = 0.1\). Also, \(\tan\theta = \frac{v^2}{Rg}\). Equating these, \(frac{v^2}{15 \times 9.8} = 0.1 ⇒ v^2 = 14.7 ⇒ v = \sqrt{14.7}\text{ m/s}\).
A road of width \(20\text{ m}\) forms an arc of radius \(15\text{ m}\), its outer edge is \(2\text{ m}\) higher than its inner edge. For what speed the road is banked?
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