## University Physics with Modern Physics (14th Edition)

The angle is $48.2^{\circ}$.
Let $R$ be the radius of the circle. Let $h$ be the change in height of the skier from the point where the skier loses contact to the top of the sphere. $\frac{1}{2}mv^2 = mgh$ $v = \sqrt{2gh}$ At the moment when the skier loses contact, the component of the weight directed toward the center of the circle is equal in magnitude to the centripetal force. $mg~cos(\alpha) = \frac{mv^2}{R}$ $g~cos(\alpha) = \frac{2gh}{R}$ $cos(\alpha) = \frac{2h}{R}$ From the diagram, we can see that $cos(\alpha) = \frac{R-h}{R}$. Therefore: $\frac{2h}{R} = \frac{R-h}{R}$ $R = 3h$ Now we can find the angle $\alpha$: $cos(\alpha) = \frac{2h}{R} = \frac{2h}{3h} = \frac{2}{3}$ $\alpha = arccos(\frac{2}{3})$ $\alpha = 48.2^{\circ}$ The angle is $48.2^{\circ}$.