A rod moves freely between the horizontal floor and the slanted wall. When the end in contact with
the floor is moving at v, what is the speed of the end in contact with the wall?
$\textbf{(A)} v\frac{\sin{\theta}}{\cos(\alpha-\theta)}$

$\textbf{(B)}v\frac{\sin(\alpha - \theta)}{\cos(\alpha + \theta)} $

$\textbf{(C)}v\frac{\cos(\alpha - \theta)}{\sin(\alpha + \theta)}$

$\textbf{(D)}v\frac{\cos(\theta)}{\cos(\alpha - \theta)}$

$\textbf{(E)}v\frac{\sin(\theta)}{\cos(\alpha + \theta)}$
1. Let's denote the length of the rod as \( L \). The rod is in contact with the horizontal floor at point \( A \) and with the slanted wall at point \( B \).
2. Let the angle between the rod and the horizontal floor be \( \theta \), and the angle between the rod and the slanted wall be \( \alpha - \theta \).
3. The velocity of point \( A \) (end in contact with the floor) is given as \( v \) in the horizontal direction.
4. Let the velocity of point \( B \) (end in contact with the wall) be \( v' \) along the direction of the incline.

Since the length of the rod remains constant, the velocities of the ends along the length of the rod must be equal. This implies that the component of the velocity of point \( A \) along the rod must be equal to the component of the velocity of point \( B \) along the rod.

5. The component of the velocity of point \( A \) along the rod is \( v \cos(\theta) \).
6. The component of the velocity of point \( B \) along the rod is \( v' \cos(\alpha - \theta) \).

Since these components must be equal, we have:
\[ v \cos(\theta) = v' \cos(\alpha - \theta) \]

7. Solving for \( v' \), we get:
\[ v' = \frac{v \cos(\theta)}{\cos(\alpha - \theta)} \]

Thus, the speed of the end in contact with the wall is:
\[ v' = v \frac{\cos(\theta)}{\cos(\alpha - \theta)} \]

The final answer is \( \boxed{ v \frac{\cos(\theta)}{\cos(\alpha - \theta)} } \).