If the Ball Never Goes Above $H$
--
In this case, **_all_** of the ball's initial kinetic energy (due to vertical motion) is converted into potential energy by the time it rises by distance $h$. Then, we have:

\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 to the potential energy gained over distance $h$ and $\Delta P$ is the potential energy gained over the distance $h$. 

If the Ball Goes Above $H$
--
The ball must go higher than $H$ to enter the basket.

In this case, **_not all_** of the ball's initial kinetic energy (due to vertical motion) is converted to potential energy by the time the ball has traveled over distance $h$. Thus $\frac{1}{2}m(v\sin{\theta})^2$ accounts for additional kinetic energy besides that which equals $mgh$.

So, we let

\begin{equation}
\begin{split}
K &= K_h + K_a &= \frac{1}{2}m(v\sin{\theta})^2 \\
P &= P_h + P_a &= (mvh) + P_a \\
K_h &= P_h \\
K_a &= P_a
\end{split}
\end{equation}

where:

* $K$ is the total kinetic energy the ball loses while going up, due to changes in the vertical component of its velocity.
* $P$ is the total potential energy the ball gains while going up.
* Subscript $_h$ refers to quantities corresponding to motion from height $L$ to height $H$, over distance $h$.
* Subscript $_a$ refers to quantities corresponding to motion at heights above $H$.

It follows that:

\begin{equation}
\begin{split}
K &= P \\
K - K_a &= P - P_a \\
K - K_a &= P_h \\
\frac{1}{2}m(v\sin{\theta})^2 - K_a &= mgh \\
(v\sin{\theta})^2 &= 2gh + \frac{2}{m}K_a\\
v\sin{\theta} &= \sqrt{2gh + \frac{2}{m}K_a}
\end{split}
\end{equation}

Since $K_a$ is non-zero in this case, we have:

\begin{equation}
\begin{split}
v\sin{\theta} &> \sqrt{2gh}
\end{split}
\end{equation}