Two boats are anchored a distance 'd' apart in a sea, where seabed is at a uniform depth of D and consists of a flat rock bed. If a gun is fired at a boat, in how much time will the firing be heard at to other boat? (Given: speed of sound in air is `v_(a)`. in the water is `v_(w)` and in the rock is `v_(r)`)

Options

A. `d//v_(a)`
B. `2D//v_(w)+d//v_(r)`
C. `(dv_(w)+2Dsqrt(v_(r)^(2)-v_(w)^(2)))/(v_(r)v_(w))`
D. `(d)/(v_(r))+((2Dv_(r))/(sqrt(v_(r)^(2)-v_(w)^(2))))xx(1)/(v_(w))`

Explanation

To solve the problem of how long it takes for the sound of a gun fired from one boat to be heard by another boat anchored a distance 'd' apart, we need to consider the different mediums through which the sound travels: air, water, and rock. Here’s a step-by-step breakdown of the solution:

### Step 1: Understand the setup
We have two boats anchored a distance 'd' apart. The seabed is at a uniform depth 'D', and consists of a flat rock bed. The sound will travel through three mediums: air, water, and rock.

### Step 2: Identify the distances
Let’s denote:
- The distance from the gun to the water surface as 'D'.
- The horizontal distance between the two boats as 'd'.
- The angle of incidence as θ.

Using trigonometry, we can find the distances traveled by sound in different mediums.

### Step 3: Calculate the distance in water
The distance traveled by sound in water (from the gun to the other boat) can be calculated using the cosine of the angle θ:
- The distance from the gun to the water surface is D.
- The horizontal distance in water (AD) can be calculated as:
\[ AD = \frac{D}{\cos(\theta)} \]

### Step 4: Calculate the distance in air
The distance traveled by sound in air (from the water surface to the other boat) can be calculated using the tangent of the angle θ:
- The horizontal distance (A) in air can be calculated as:
\[ A = D \tan(\theta) \]

### Step 5: Total time calculation
The total time \( T \) for the sound to travel from the gun to the other boat can be expressed as:
\[ T = T_{AD} + T_{BC} + T_{CD} \]
Where:
- \( T_{AD} \) is the time taken for sound to travel in water.
- \( T_{BC} \) is the time taken for sound to travel in air.
- \( T_{CD} \) is the time taken for sound to travel in rock.

### Step 6: Express time in terms of distance and speed
Using the formula \( T = \frac{Distance}{Speed} \), we can express each time component:
- For sound in water:
\[ T_{AD} = \frac{2D}{v_w} \]
- For sound in air:
\[ T_{BC} = \frac{d - 2A}{v_a} \]
Where \( A = D \tan(\theta) \).

### Step 7: Combine the times
Now, substituting the expressions for \( T_{AD} \) and \( T_{BC} \) into the total time equation:
\[ T = 2 \left(\frac{D}{v_w \cos(\theta)}\right) + \frac{d - 2D \tan(\theta)}{v_a} \]

### Step 8: Differentiate to find the optimal angle
To find the optimal angle θ for minimum time, we differentiate \( T \) with respect to θ and set it to zero:
\[ \frac{dT}{d\theta} = 0 \]

### Step 9: Solve for θ
From the differentiation, we can derive that:
\[ \sin(\theta) = \frac{v_w}{v_a} \]
And we can find \( \tan(\theta) \) as:
\[ \tan(\theta) = \frac{v_w}{\sqrt{v_a^2 - v_w^2}} \]

### Step 10: Substitute back to find total time
Finally, substitute the values of \( \tan(\theta) \) and \( \cos(\theta) \) back into the total time equation to find the final expression for \( T \).

### Final Expression
The final expression for the time \( T \) will be:
\[ T = \frac{2D}{v_w \cos(\theta)} + \frac{d - 2D \tan(\theta)}{v_a} \]