A spring with natural length h and spring constant k has one end...

Let the initial length of the spring be $L$. From the given geometry, the vertical height is $h$ and the spring makes an angle of $37^\circ$ with the vertical.

$L \cos 37^\circ = h$

$L = \dfrac{h}{\cos 37^\circ} = \dfrac{h}{4/5} = \dfrac{5h}{4}$

The initial extension in the spring is:

$x = L - h = \dfrac{5h}{4} - h = \dfrac{h}{4}$

The initial mechanical energy of the system consists only of the elastic potential energy of the spring (since the ring is released from rest and gravitational potential energy remains constant):

$E_i = \dfrac{1}{2}kx^2 = \dfrac{1}{2}k\left(\dfrac{h}{4}\right)^2 = \dfrac{kh^2}{32}$

When the spring becomes vertical, its length is equal to the height $h$, which is its natural length. Thus, the final elastic potential energy is zero. The final mechanical energy is purely the kinetic energy of the ring:

$E_f = \dfrac{1}{2}mv^2$

By the principle of conservation of mechanical energy:

$E_i = E_f$

$\dfrac{1}{2}mv^2 = \dfrac{kh^2}{32}$

$v^2 = \dfrac{kh^2}{16m}$

$v = \dfrac{h}{4}\sqrt{\dfrac{k}{m}}$