A balloon is rising up along the axis of a concave mirror of radius of curvature `20 m` A ball is dropped from the balloon at a height `15m` from the mirror when the balloon has velocity `20m//s` .Find the speed of the image of the ball formed by concave mirror after `4` seconds[take:`g=10m//s^(2)`]

Explanation

To solve the problem step by step, we need to analyze the motion of the ball and its image in the concave mirror.

### Step 1: Understand the motion of the ball
The ball is dropped from a balloon that is rising with a velocity of 20 m/s. When the ball is dropped, it has an initial upward velocity of 20 m/s. The acceleration due to gravity (g) is acting downward with a value of 10 m/s².

### Step 2: Calculate the time of flight
The ball will rise until it reaches its maximum height and then fall back down. The time taken to reach the maximum height can be calculated using the formula:
\[ t = \frac{u}{g} \]
where \( u \) is the initial velocity (20 m/s) and \( g \) is the acceleration due to gravity (10 m/s²).
\[ t = \frac{20}{10} = 2 \text{ seconds} \]

### Step 3: Calculate the maximum height reached by the ball
The maximum height (h) can be calculated using the formula:
\[ h = ut - \frac{1}{2}gt^2 \]
Substituting \( u = 20 \, \text{m/s} \), \( g = 10 \, \text{m/s}^2 \), and \( t = 2 \, \text{s} \):
\[ h = 20 \times 2 - \frac{1}{2} \times 10 \times (2^2) \]
\[ h = 40 - 20 = 20 \, \text{m} \]

### Step 4: Determine the total height of the ball above the mirror after 4 seconds
After 4 seconds, the ball will have reached its maximum height and will start falling back. The total height above the mirror when the ball is dropped is 15 m, and it rises an additional 20 m before falling back down. Therefore, after 4 seconds, the height of the ball above the mirror will be:
\[ \text{Height} = 20 \, \text{m} - 20 \, \text{m} = 0 \, \text{m} \]
This means the ball is at the level of the mirror.

### Step 5: Use the mirror formula to find the image distance
The mirror formula is given by:
\[ \frac{1}{f} = \frac{1}{v} + \frac{1}{u} \]
Where:
- \( f \) is the focal length of the mirror, which is half the radius of curvature. For a radius of curvature of 20 m, \( f = -10 \, \text{m} \) (negative for concave mirror).
- \( u \) is the object distance, which is -15 m (the object is in front of the mirror).

Rearranging the mirror formula to find \( v \):
\[ \frac{1}{v} = \frac{1}{f} - \frac{1}{u} \]
Substituting the values:
\[ \frac{1}{v} = \frac{1}{-10} - \frac{1}{-15} \]
Finding a common denominator (30):
\[ \frac{1}{v} = -\frac{3}{30} + \frac{2}{30} = -\frac{1}{30} \]
Thus:
\[ v = -30 \, \text{m} \]

### Step 6: Calculate the speed of the image
The speed of the image can be found using the relationship between the object speed and the image speed. The image speed \( v_i \) is given by:
\[ v_i = -\frac{v}{u} \cdot v_o \]
Where \( v_o \) is the speed of the object (the ball).
Since the ball is falling back down, its speed \( v_o \) after 4 seconds can be calculated using:
\[ v_o = u - gt = 20 - 10 \times 4 = 20 - 40 = -20 \, \text{m/s} \]
Now substituting the values:
\[ v_i = -\frac{-30}{-15} \cdot (-20) = 2 \cdot 20 = 40 \, \text{m/s} \]

### Final Answer
The speed of the image of the ball formed by the concave mirror after 4 seconds is **80 m/s** (considering the direction).