Let $y$ be the vertical height of the hand and $x$ be the horizontal position of the block. The length of the string is $l$, so $y = l \sin \theta$. The vertical velocity of the hand is constant, $v_0$. $$ v_0 = \frac{dy}{dt} = \frac{d}{dt}(l \sin \theta) = l \cos \theta \cdot \dot{\theta} $$ $$ \implies \dot{\theta} = \frac{v_0}{l \cos \theta} $$

The horizontal position $x = l \cos \theta$. The block’s acceleration $a_x$ is $\ddot{x}$. $$ v_x = \dot{x} = -l \sin \theta \cdot \dot{\theta} $$ $$ a_x = \frac{d v_x}{dt} = -l [\cos \theta \cdot (\dot{\theta})^2 + \sin \theta \cdot \ddot{\theta}] $$ Alternatively, relating directly to $v_0$:
From $x^2 + y^2 = l^2 \implies x v_x + y v_0 = 0 \implies v_x = -v_0 \tan \theta$. $$ a_x = \frac{d}{dt}(-v_0 \tan \theta) = -v_0 \sec^2 \theta \cdot \dot{\theta} $$ Substitute $\dot{\theta}$: $$ |a_x| = v_0 \sec^2 \theta \left( \frac{v_0}{l \cos \theta} \right) = \frac{v_0^2}{l \cos^3 \theta} $$

The block leaves the floor when the Normal force becomes zero. Vertical forces: $T \sin \theta = mg \implies T = \frac{mg}{\sin \theta}$.

Horizontal forces provide the acceleration: $$ T \cos \theta = m |a_x| $$ Substituting $T$ and $|a_x|$: $$ \left( \frac{mg}{\sin \theta} \right) \cos \theta = m \left( \frac{v_0^2}{l \cos^3 \theta} \right) $$

Simplify the equation: $$ g \cot \theta = \frac{v_0^2}{l \cos^3 \theta} $$ $$ v_0^2 = gl \cot \theta \cos^3 \theta = gl \frac{\cos^4 \theta}{\sin \theta} $$ $$ v_0 = \cos^2 \theta \sqrt{\frac{gl}{\sin \theta}} $$