Here is a concise solution focusing on key concepts and essential derivation steps.

1. Kinematics of the Stick's Center

Let the corner between the wall and the ground be the origin. The position of the stick's ends are $A(0, y_A)$ and $B(x_B, 0)$. The stick has length $2l$, so $x_B^2 + y_A^2 = (2l)^2$. Let $\theta$ be the angle the stick makes with the ground. The coordinates of the ends are $x_B = 2l \cos\theta$ and $y_A = 2l \sin\theta$. The ball is at the center of the stick, $C(x_c, y_c)$, where:

$$x_c = l \cos\theta$$ $$y_c = l \sin\theta$$

2. Acceleration of the Ball

The end B moves at a constant horizontal velocity $v$, so $v_B = \dot{x}_B = v$, and its acceleration is $a_B = \ddot{x}_B = 0$.

From the geometric relation $x_c = x_B/2$, we can find the horizontal acceleration of the ball by differentiating twice with respect to time:

$$\dot{x}_c = \frac{\dot{x}_B}{2} = \frac{v}{2}$$ $$a_{cx} = \ddot{x}_c = \frac{\ddot{x}_B}{2} = 0$$

The horizontal acceleration of the ball is zero.

To find the vertical acceleration, we first relate the angular velocity $\dot{\theta}$ to $v$:

$$v = \dot{x}_B = \frac{d}{dt}(2l \cos\theta) = -2l \sin\theta \, \dot{\theta}$$ $$\dot{\theta} = -\frac{v}{2l \sin\theta}$$

Now, we find the vertical acceleration $a_{cy}$ by differentiating $y_c$ twice:

$$v_{cy} = \dot{y}_c = \frac{d}{dt}(l \sin\theta) = l \cos\theta \, \dot{\theta}$$

Substituting $\dot{\theta}$:

$$v_{cy} = l \cos\theta \left(-\frac{v}{2l \sin\theta}\right) = -\frac{v}{2} \cot\theta$$

Differentiating again to find acceleration:

$$a_{cy} = \dot{v}_{cy} = \frac{d}{dt}\left(-\frac{v}{2} \cot\theta\right) = -\frac{v}{2} (-\csc^2\theta) \dot{\theta} = \frac{v}{2 \sin^2\theta} \dot{\theta}$$

Substituting $\dot{\theta}$ again:

$$a_{cy} = \frac{v}{2 \sin^2\theta} \left(-\frac{v}{2l \sin\theta}\right) = -\frac{v^2}{4l \sin^3\theta}$$

At the specified angle $\theta = 45^\circ$, we have $\sin(45^\circ) = 1/\sqrt{2}$:

$$a_{cy} = -\frac{v^2}{4l (1/\sqrt{2})^3} = -\frac{v^2}{4l (1/2\sqrt{2})} = -\frac{\sqrt{2}v^2}{2l}$$

The acceleration of the ball is $\vec{a}_c = a_{cx}\hat{i} + a_{cy}\hat{j} = -\frac{\sqrt{2}v^2}{2l} \hat{j}$.

3. Force on the Stick

The forces acting on the ball are gravity $\vec{F}_g = -mg\hat{j}$ and the force from the stick, $\vec{F}_{stick \to ball}$. Applying Newton's second law to the ball:

$$\vec{F}_{stick \to ball} + \vec{F}_g = m\vec{a}_c$$ $$\vec{F}_{stick \to ball} = m\vec{a}_c - \vec{F}_g = m\left(-\frac{\sqrt{2}v^2}{2l}\hat{j}\right) - (-mg\hat{j})$$ $$\vec{F}_{stick \to ball} = \left(mg - \frac{m\sqrt{2}v^2}{2l}\right)\hat{j}$$

The problem asks for the force of the ball applied on the stick, $\vec{F}_{ball \to stick}$. By Newton's third law, this is the reaction force:

$$\vec{F}_{ball \to stick} = -\vec{F}_{stick \to ball}$$ $$\vec{F}_{ball \to stick} = -\left(mg - \frac{m\sqrt{2}v^2}{2l}\right)\hat{j} = \left(\frac{m\sqrt{2}v^2}{2l} - mg\right)\hat{j}$$

This force is purely vertical. A non-calculus method is not feasible as calculating acceleration from a non-uniform motion inherently requires differentiation.