A stone is dropped from the top of a tower of 125 m high into a pond which is at base of the tower. When will the splash be heard at the top? (Given g - 10 m `s^(-2)` and speed of sound = 340 m `s^(-1)`.)

Options

A. 5s
B. 0.36 s
C. 5.36s
D. 2s

Explanation

To solve the problem of when the splash will be heard at the top of a 125 m tower after a stone is dropped, we need to calculate two times: the time it takes for the stone to fall (t₁) and the time it takes for the sound of the splash to travel back up to the top of the tower (t₂).

### Step-by-Step Solution:

1. **Calculate the time taken for the stone to fall (t₁)**:
- We can use the equation of motion:
\[ s = ut + \frac{1}{2}gt^2 \]
- Here, \( s \) is the distance (125 m), \( u \) is the initial velocity (0 m/s, since the stone is dropped), \( g \) is the acceleration due to gravity (10 m/s²), and \( t \) is the time taken to fall (t₁).
- Substituting the known values:
\[ 125 = 0 \cdot t₁ + \frac{1}{2} \cdot 10 \cdot t₁^2 \]
- This simplifies to:
\[ 125 = 5t₁^2 \]
- Rearranging gives:
\[ t₁^2 = \frac{125}{5} = 25 \]
- Taking the square root:
\[ t₁ = 5 \text{ seconds} \]

2. **Calculate the time taken for the sound to travel back up (t₂)**:
- The speed of sound is given as 340 m/s. The distance the sound travels is also 125 m.
- We can use the formula:
\[ t₂ = \frac{\text{distance}}{\text{speed}} = \frac{125}{340} \]
- Calculating this gives:
\[ t₂ \approx 0.368 \text{ seconds} \quad (\text{approximately } 0.36 \text{ seconds}) \]

3. **Calculate the total time (T)**:
- The total time to hear the splash at the top of the tower is the sum of the two times:
\[ T = t₁ + t₂ = 5 + 0.36 = 5.36 \text{ seconds} \]

### Final Answer:
The splash will be heard at the top of the tower after approximately **5.36 seconds**.

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