A stone is released from the top of a tower of height 19.6m. Calculate its final velocity just before touching the ground.

Explanation

To solve the problem of calculating the final velocity of a stone just before it touches the ground after being released from a height of 19.6 meters, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the known values**:
- Height (s) = 19.6 m
- Initial velocity (u) = 0 m/s (since the stone is released)
- Acceleration due to gravity (g) = 9.8 m/s²

2. **Use the third equation of motion**:
The third equation of motion states:
\[ v^2 = u^2 + 2gs \]
where:
- \( v \) is the final velocity,
- \( u \) is the initial velocity,
- \( g \) is the acceleration due to gravity,
- \( s \) is the distance fallen (height of the tower).

3. **Substitute the known values into the equation**:
Since \( u = 0 \):
\[ v^2 = 0^2 + 2 \cdot 9.8 \cdot 19.6 \]
Simplifying this gives:
\[ v^2 = 2 \cdot 9.8 \cdot 19.6 \]

4. **Calculate the value**:
\[ v^2 = 2 \cdot 9.8 \cdot 19.6 = 384.16 \]

5. **Find the final velocity (v)**:
To find \( v \), take the square root of \( v^2 \):
\[ v = \sqrt{384.16} \approx 19.6 \text{ m/s} \]

### Final Answer:
The final velocity of the stone just before it touches the ground is approximately **19.6 m/s**.