A man standing in front of a mountain beats a drum at regular intervals. The drumming rate is gradually increased and he finds that echo is not heard distinctly when the rate becomes `40` per minute. He then moves near to the mountain by `90` metres and finds that echo is again not heard distinctly when the drumming rate becomes `60` per minute. Calculate (a) the distance between the mountain and the initial position of the man and (b) the velocity of sound.

Explanation

To solve the problem step by step, let's break it down into manageable parts.

### Step 1: Understand the Problem
The man beats a drum at a rate of 40 beats per minute and does not hear the echo distinctly. When he moves 90 meters closer to the mountain, the rate increases to 60 beats per minute, and he still does not hear the echo distinctly. We need to find the initial distance to the mountain and the speed of sound.

### Step 2: Convert Beats per Minute to Time Interval
First, we need to convert the drumming rates into time intervals.

- For 40 beats per minute:
\[ \text{Time interval} (T_1) = \frac{60 \text{ seconds}}{40 \text{ beats}} = 1.5 \text{ seconds} \]

- For 60 beats per minute:
\[ \text{Time interval} (T_2) = \frac{60 \text{ seconds}}{60 \text{ beats}} = 1 \text{ second} \]

### Step 3: Relate Time Intervals to Distances
The time taken for the echo to return is related to the distance to the mountain and the speed of sound (v). The formula for the time taken for the echo to return is:
\[ T = \frac{2s}{v} \]
where \( s \) is the distance to the mountain.

- For the initial position:
\[ 1.5 = \frac{2s}{v} \quad \text{(1)} \]

- For the position after moving 90 meters closer:
\[ 1 = \frac{2(s - 90)}{v} \quad \text{(2)} \]

### Step 4: Solve the Equations
From equation (1):
\[ v = \frac{2s}{1.5} = \frac{4s}{3} \quad \text{(3)} \]

Substituting equation (3) into equation (2):
\[ 1 = \frac{2(s - 90)}{\frac{4s}{3}} \]
Cross-multiplying gives:
\[ \frac{4s}{3} = 2(s - 90) \]
Simplifying this:
\[ 4s = 6(s - 90) \]
\[ 4s = 6s - 540 \]
Rearranging gives:
\[ 2s = 540 \quad \Rightarrow \quad s = 270 \text{ meters} \]

### Step 5: Calculate the Velocity of Sound
Now, substituting \( s = 270 \) meters back into equation (3):
\[ v = \frac{4 \times 270}{3} = 360 \text{ m/s} \]

### Final Answers
(a) The distance between the mountain and the initial position of the man is **270 meters**.
(b) The velocity of sound is **360 m/s**.