A ball is dropped from a height of `20m` above the surface of water in a lake. The refractive index of water is `4//3`. A fish inside the lake, in the line of fall of the ball, is looking at the ball. At an instant, when the ball is `12.8m` above the water surface, the fish sees the speed of ball as `(g=10m//sec^2)`

Options

A. `9m//sec`
B. `12m//sec`
C. `16m//sec`
D. `21.33m//sec`

Explanation

To solve the problem step by step, we will follow these calculations:

### Step 1: Determine the height dropped by the ball
The ball is dropped from a height of 20 m and at the instant of interest, it is 12.8 m above the water surface.

\[ \text{Height dropped} = 20 \, \text{m} - 12.8 \, \text{m} = 7.2 \, \text{m} \]

### Step 2: Calculate the speed of the ball just before it reaches 12.8 m
Using the equation of motion under gravity, we can find the speed of the ball just before it reaches the height of 12.8 m. The formula used is:

\[ v^2 = u^2 + 2gh \]

Where:
- \(u = 0\) (initial speed, since it is dropped)
- \(g = 10 \, \text{m/s}^2\) (acceleration due to gravity)
- \(h = 7.2 \, \text{m}\) (height fallen)

Substituting the values:

\[ v^2 = 0 + 2 \times 10 \times 7.2 \]
\[ v^2 = 144 \]
\[ v = \sqrt{144} = 12 \, \text{m/s} \]

### Step 3: Calculate the apparent speed of the ball as seen by the fish
The fish is in water, and the refractive index of water is given as \( \mu = \frac{4}{3} \). The apparent speed of the ball as seen by the fish is given by:

\[ v' = \mu v \]

Where:
- \(v = 12 \, \text{m/s}\) (actual speed of the ball)
- \(\mu = \frac{4}{3}\)

Substituting the values:

\[ v' = \frac{4}{3} \times 12 \]
\[ v' = \frac{48}{3} = 16 \, \text{m/s} \]

### Final Answer
The speed of the ball as seen by the fish is \(16 \, \text{m/s}\).

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