A marble rolls on a horizontal table with a speed $v_0$. The marble bounces elastically in a vertical cut at a horizontal distance $D$ from the edge of the table. After hitting the ground the marble will record a $x_0$ distance to the edge of the table.

$a)$ Determine the equations for horizontal and vertical motion in function of time before hitting the wall.
$b)$ Determine the height of the marble at the time of the collision.
$c)$ Determine the vertical speed at the instant of the collision.
$d)$ Determine the value of $v_0$ for which $x_0 = 0$

[i]Proposed by Danilo Molina, Colon[/i]
### Part (a)
Determine the equations for horizontal and vertical motion as functions of time before hitting the wall.

1. **Horizontal Motion:**
   The horizontal motion is uniform because there is no horizontal acceleration. The equation for horizontal displacement \( x(t) \) is:
   \[
   x(t) = v_0 t
   \]
   where \( v_0 \) is the initial horizontal speed.

2. **Vertical Motion:**
   The vertical motion is uniformly accelerated motion due to gravity. The equation for vertical displacement \( y(t) \) is:
   \[
   y(t) = h - \frac{1}{2} g t^2
   \]
   where \( h \) is the initial height and \( g \) is the acceleration due to gravity.

### Part (b)
Determine the height of the marble at the time of the collision.

1. The time of collision \( t_c \) is when the marble reaches the wall at distance \( D \):
   \[
   t_c = \frac{D}{v_0}
   \]

2. Substitute \( t_c \) into the vertical motion equation to find the height at the time of collision:
   \[
   y(t_c) = h - \frac{1}{2} g \left( \frac{D}{v_0} \right)^2
   \]
   Simplifying, we get:
   \[
   y(t_c) = h - \frac{g D^2}{2 v_0^2}
   \]

### Part (c)
Determine the vertical speed at the instant of the collision.

1. The vertical speed \( v_y(t) \) is given by the derivative of the vertical displacement \( y(t) \):
   \[
   v_y(t) = \frac{d}{dt} \left( h - \frac{1}{2} g t^2 \right) = -g t
   \]

2. At the time of collision \( t_c \):
   \[
   v_y(t_c) = -g \left( \frac{D}{v_0} \right)
   \]
   Simplifying, we get:
   \[
   v_y(t_c) = -\frac{g D}{v_0}
   \]

### Part (d)
Determine the value of \( v_0 \) for which \( x_0 = 0 \).

1. If \( x_0 = 0 \), the marble must return to the edge of the table after bouncing off the wall. This implies that the marble travels a total horizontal distance of \( 2D \) before hitting the ground.

2. The time \( t_{air} \) for the marble to fall from height \( h \) to the ground is given by:
   \[
   h = \frac{1}{2} g t_{air}^2 \implies t_{air} = \sqrt{\frac{2h}{g}}
   \]

3. The horizontal distance traveled in this time must be \( 2D \):
   \[
   2D = v_0 t_{air}
   \]

4. Substitute \( t_{air} \):
   \[
   2D = v_0 \sqrt{\frac{2h}{g}}
   \]

5. Solve for \( v_0 \):
   \[
   v_0 = \frac{2D}{\sqrt{\frac{2h}{g}}} = 2D \sqrt{\frac{g}{2h}}
   \]

The final answer is \( \boxed{ v_0 = 2D \sqrt{\frac{g}{2h}} } \)