A ball is dropped from rest 3.0 m directly above the vertex of a concave mirror that has a radius
of 1.0 m and lies in a horizontal plane.
(a) Describe the motion of ball's image in the mirror
(b) At what time do the ball and its image coincide?

Explanation

To solve the problem step by step, we will address both parts of the question regarding the motion of the ball's image in the concave mirror and the time at which the ball and its image coincide.

### Step-by-Step Solution:

**Given:**
- Height from which the ball is dropped (h) = 3.0 m
- Radius of the concave mirror (R) = 1.0 m
- Focal length (f) of the concave mirror = -R/2 = -0.5 m (negative because it is a concave mirror)

### Part (a): Describe the motion of the ball's image in the mirror

1. **Determine the position of the ball:**
The ball is dropped from a height of 3.0 m above the mirror. The distance from the mirror to the ball at any time \( t \) can be calculated using the equation of motion:
\[ h(t) = h_0 - \frac{1}{2} g t^2 \]
where \( h_0 = 3.0 \, \text{m} \) and \( g = 9.8 \, \text{m/s}^2 \).

2. **Calculate the position of the image:**
The mirror formula is given by:
\[ \frac{1}{f} = \frac{1}{u} + \frac{1}{v} \]
where \( u \) is the object distance (distance of the ball from the mirror) and \( v \) is the image distance. Since the ball is falling, \( u \) will change with time.

3. **Determine the nature of the image:**
- When the ball is above the focal point (between the focal point and the mirror), the image will be real and inverted.
- When the ball is between the focal point and the mirror, the image will be virtual and upright.

4. **Motion of the image:**
As the ball falls, the image will move according to the mirror formula, and its position will change based on the changing object distance \( u \). The image will initially be real and will become virtual as the ball passes the focal point.

### Part (b): At what time do the ball and its image coincide?

1. **Calculate the time for the ball to fall to the mirror:**
The distance the ball falls to the mirror is 3.0 m:
\[ s_1 = 2.0 \, \text{m} \quad (\text{from 3.0 m to 1.0 m}) \]
Using the equation of motion:
\[ s_1 = \frac{1}{2} g t_1^2 \]
Rearranging gives:
\[ t_1 = \sqrt{\frac{2s_1}{g}} = \sqrt{\frac{2 \times 2.0}{9.8}} \approx 0.639 \, \text{s} \]

2. **Calculate the time for the image to reach the same height:**
The distance the image travels is from the focal point to the center of curvature (which is 1.0 m):
\[ s_2 = 3.0 \, \text{m} \quad (\text{from the focal point to the mirror}) \]
Using the same equation of motion:
\[ t_2 = \sqrt{\frac{2s_2}{g}} = \sqrt{\frac{2 \times 3.0}{9.8}} \approx 0.785 \, \text{s} \]

3. **Determine when they coincide:**
The ball and its image will coincide when they have traveled the same vertical distance. This occurs when both times \( t_1 \) and \( t_2 \) are equal.

### Final Answer:
- The ball and its image coincide approximately at \( t \approx 0.785 \, \text{s} \).