There seems to be a much simpler way to go about this.
TL;DNR Version
\begin{equation} \begin{split} \Delta K &= \Delta P \\ \frac{1}{2}m(v\sin{\theta})^2 &= mgh \\ v\sin{\theta} &= \sqrt{2gh} \end{split} \end{equation}
Where $\Delta K$ is the kinetic energy lost over distance $h$ and $\Delta P$ is the potential energy gained over the distance $h$.
Longer Explanation
Consider qualitatively:
- The initial velocity of the ball determines the initial value for $K$, the ball's kinetic energy.
- As the ball rises higher, some of that initial kinetic energy is converted into potential energy, $P$.
- Total energy $K$ + $P$ is preserved. That is, $$ \begin{equation} K_0 + P_0 = K_b + P_b \end{equation} $$
Where subscript $_0$ refers to values at the beginning of the experiment and subscript $_b$ refers to values at $t_b$, as you have defined it. Now, suppose that: $$ \begin{equation} v_x = v\cos(\theta) \\ v_y = v\sin(\theta) \end{equation} $$ We know that: \begin{equation} \begin{split} K & = \frac{1}{2}mv^2 \\ & = \frac{1}{2}m(\sqrt{x_x^2+v_y^2})^2 \\ & = \frac{1}{2}mv_x^2 + \frac{1}{2}mv_y^2 \end{split} \end{equation} Thus, accelerating the ball in the direction of $v_y$ has an additive effect on $K$, regardless of what $v_x$ is doing. Same rule applies to $v_x$, regardless of $v_y$.
Since $g$ operates exclusively in the direction of $v_y$ and there is no air resistance, we can conclude that $v_x$ does not change during the part of the experiment we are interested in. Thus, we can exclude $v_x$ from any calculations which add or remove energy from $K$ as long as we are only considering changes which occur during the part of the experiment we are interested in.
So, how much energy is added or removed from $K$? Since the ball is moving slower at the top of its arc than at the bottom, we can conclude that it loses $K$ as it rises. Specifically, it would lose $K_0 - K_b$. It then follows from conservation of energy that: $$ \begin{equation} \begin{split} K_0 + P_0 &= K_b + P_b \\ K_0 - K_b &= P_b - P_0 \end{split} \end{equation} $$
In other words, $P$ gains everything that $K$ loses. So what does $P$ gain? We know that:
- The ball rises in the $v_y$ direction by distance ($\Delta y = h$).
- Over that distance, also in the $v_y$ direction, it experienced a constant force $F = mg$.
We know from the definition of potential energy that: $$ \Delta P = F\Delta y \\ $$ So, in this case: $$ P_b - P_0 = mgh $$
It then follows that: $$ \begin{equation} \begin{split} K_0 - K_b &= mgh \\ \frac{1}{2}mv_y^2 &= mgh \\ v_y &= \sqrt{2gh} \\ v\sin{\theta} &= \sqrt{2gh} \end{split} \end{equation} $$
Note that the above equations only represent energy converted over the distance $h$. If the ball rose above $h$ before falling back down into the hoop, the potential energy gained above $h$ would be lost when the ball fell back down to $h$. Conversely, the kinetic energy lost above $h$ would be regained when the ball fell back down to $h$. Thus, the net change in both $K$ and $P$ for all travel above $h$ during the experiment would be $0$.