A fruit which is at 20 m above a lake suddenly detaches from the tree.A fish inside the lake , in the line of fall of fruit,is looking at fruit when fruit is at 12.8 m above the water surface.Speed of fruit as observed by fish `(μ_(water)=4/3 and g=10m/s^2)`

Options

A. 6 m/s
B. 12 m/s
C. 15 m/s
D. 16 m/s

Answer

To solve the problem step by step, we will follow the physics principles involved in the motion of the fruit and the effect of the refractive index on the speed observed by the fish.

### Step 1: Determine the displacement of the fruit
The fruit is initially at a height of 20 m and is at a height of 12.8 m when observed by the fish. The displacement \( s \) of the fruit is given by:
\[ s = 20 \, \text{m} - 12.8 \, \text{m} = 7.2 \, \text{m} \]

### Step 2: Use the kinematic equation to find the speed of the fruit
Since the fruit is dropped from rest, its initial velocity \( u = 0 \) m/s. We can use the kinematic equation:
\[ v^2 = u^2 + 2as \]
where:
- \( v \) is the final velocity,
- \( u \) is the initial velocity (0 m/s),
- \( a \) is the acceleration due to gravity (given as \( g = 10 \, \text{m/s}^2 \)),
- \( s \) is the displacement (7.2 m).

Substituting the values:
\[ v^2 = 0 + 2 \cdot 10 \cdot 7.2 \]
\[ v^2 = 144 \]
\[ v = \sqrt{144} = 12 \, \text{m/s} \]

### Step 3: Calculate the speed of the fruit as observed by the fish
The speed of the fruit as observed by the fish in the water is affected by the refractive index of water. The relationship is given by:
\[ V_f = \mu \cdot v \]
where:
- \( V_f \) is the speed of the fruit as observed by the fish,
- \( \mu \) is the refractive index of water (given as \( \frac{4}{3} \)),
- \( v \) is the speed of the fruit in air (12 m/s).

Substituting the values:
\[ V_f = \frac{4}{3} \cdot 12 \]
\[ V_f = \frac{48}{3} = 16 \, \text{m/s} \]

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

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