A toboggan travels along the path $ABC$ shown in the diagram. The path lies in a vertical plane, and consists of two circular arcs $AB$ and $BC$. The arcs are of radius 20 m, and subtend angles of $60^{\circ}$ at their centres, $D$ and $E$ respectively. The line $ABC$ is horizontal, and there is no friction between the toboggan and the snow. Air resistance is negligible, and the toboggan may be treated as a particle. The speed of the toboggan at its lowest point is $U \mathrm{ms^{-1}}$.
Find the range of values of $U$ for which the toboggan will reach $C$ without losing contact with the snow.
Find also the value of $U$ for which the toboggan will leave the snow at $B$, and travel as a projectile until it lands again at $C$.
My thoughts:
I know that toboggan will gain potential energy and lose kninetic energy once it rises to a small height of about $2.5 \mathrm{m}$ to reach $C$ with a zero speed and plug these values in energy equation assuming that this is lowest possible value of $U$ and unfortunately the answer is incorrect so how to find extremities of inequality of $U$. Answers are: $10.4 < U < 15.1; 16.9$
The first part. Let's calulate the mínimun velocity for the body to reach the higher part of the circular path. Consider the conservation of mechanical energy
Some trigonometry leads to the difference in heigh $\Delta h=40\left(1-\frac{\sqrt{3}}{2}\right)$
For the mechanical energy to be conserved $\frac{1}{2}mU_{min}^2=mg\Delta h$
$$U_{min}=\sqrt{80g\left(1-\frac{\sqrt{3}}{2}\right)}$$
The maximum velocity is such that in B the body doesn't "touch" it, so is, the gravity force is enough to give the body it's circular trajectory at that point. This is so because for parabolic trajectory, the curvature is the lesser the farther is the point from the vertex and with such a condition is for sure that the body never loses contact ($R=20$):
$$m\frac{v^2}{R}=mg\frac{\sqrt{3}}{2}$$
Being the left hand side the centripetal acceleration times the mass, $v$ the velocity at that point.
Mechanical energy is conserved, so
$$\frac{1}{2}mU_{max}^2=\frac{1}{2}mv^2+\frac{1}{2}mg\Delta h$$
$$\frac{1}{2}mU_{max}^2=\frac{1}{2}mgR\frac{\sqrt{3}}{2}+\frac{1}{2}mg\Delta h$$
$$U_{max}=\sqrt{g\left(R\frac{\sqrt{3}}{2}+\Delta h\right)}$$
For the second part you need to consider a parabolic flight starting at B and reachig C.