Answer
$$\theta=9.20 \mathrm{\ m} $$
Work Step by Step
the potential energy at the time shown is
$$
U=-m g R(1-\cos \theta)
$$
He starts from rest and his kinetic energy at the time shown is $\frac{1}{2} m v^{2} .$ Thus conservation
of energy gives
$$
0=\frac{1}{2} m v^{2}-m g R(1-\cos \theta)
$$
or $$v^{2}=2 g R(1-\cos \theta) .$$ We substitute this expression into the equation developed from the second law to obtain $$g \cos \theta=2 g(1-\cos \theta) .$$ This gives $$\cos \theta=2 / 3 .$$ The height of the boy above the bottom of the mound is $$\theta=\frac{2}{3} R=\frac{2}{3}(13.8 \mathrm{m})=9.20 \mathrm{\ m} $$
