A man releases a stone at the top edge of a tower. During the last second of its travel, the stone falls through a distance of (9/15) H, where H is the tower's height. Find H.

Answer

To solve the problem, we need to find the height \( H \) of the tower given that a stone falls through a distance of \( \frac{9}{15} H \) during the last second of its travel.

### Step-by-Step Solution:

1. **Understanding the Motion**:
- The stone is released from rest, so the initial velocity \( u = 0 \).
- The stone falls under the influence of gravity, with acceleration \( g \approx 9.81 \, \text{m/s}^2 \).

2. **Distance Fallen in \( t \) Seconds**:
- The distance fallen after \( t \) seconds can be calculated using the second equation of motion:
\[ S_t = ut + \frac{1}{2} g t^2 \]
- Since \( u = 0 \), this simplifies to:
\[ S_t = \frac{1}{2} g t^2 \]
- Therefore, the total distance fallen in \( t \) seconds is:
\[ H = \frac{1}{2} g t^2 \]

3. **Distance Fallen in \( t-1 \) Seconds**:
- The distance fallen in \( t-1 \) seconds is:
\[ S_{t-1} = \frac{1}{2} g (t-1)^2 \]

4. **Distance Fallen in the Last Second**:
- The distance fallen during the last second (from \( t-1 \) to \( t \)) is given by:
\[ S_{\text{last}} = S_t - S_{t-1} \]
- Substituting the expressions for \( S_t \) and \( S_{t-1} \):
\[ S_{\text{last}} = \frac{1}{2} g t^2 - \frac{1}{2} g (t-1)^2 \]

5. **Simplifying the Distance Fallen in the Last Second**:
- Expanding \( (t-1)^2 \):
\[ (t-1)^2 = t^2 - 2t + 1 \]
- Thus,
\[ S_{\text{last}} = \frac{1}{2} g t^2 - \frac{1}{2} g (t^2 - 2t + 1) = \frac{1}{2} g (2t - 1) \]

6. **Setting Up the Equation**:
- We know from the problem statement that:
\[ S_{\text{last}} = \frac{9}{15} H = \frac{3}{5} H \]
- Therefore, we can equate:
\[ \frac{1}{2} g (2t - 1) = \frac{3}{5} H \]

7. **Substituting for \( H \)**:
- From the expression for \( H \):
\[ H = \frac{1}{2} g t^2 \]
- Substituting this into the equation gives:
\[ \frac{1}{2} g (2t - 1) = \frac{3}{5} \left(\frac{1}{2} g t^2\right) \]
- Canceling \( \frac{1}{2} g \) from both sides (assuming \( g \neq 0 \)):
\[ 2t - 1 = \frac{3}{5} t^2 \]

8. **Rearranging the Equation**:
- Rearranging gives:
\[ \frac{3}{5} t^2 - 2t + 1 = 0 \]
- Multiplying through by 5 to eliminate the fraction:
\[ 3t^2 - 10t + 5 = 0 \]

9. **Using the Quadratic Formula**:
- The quadratic formula is given by:
\[ t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
- Here, \( a = 3 \), \( b = -10 \), and \( c = 5 \):
\[ t = \frac{10 \pm \sqrt{(-10)^2 - 4 \cdot 3 \cdot 5}}{2 \cdot 3} = \frac{10 \pm \sqrt{100 - 60}}{6} = \frac{10 \pm \sqrt{40}}{6} = \frac{10 \pm 2\sqrt{10}}{6} = \frac{5 \pm \sqrt{10}}{3} \]

10. **Finding \( H \)**:
- We can use \( t \) to find \( H \):
\[ H = \frac{1}{2} g t^2 \]
- Substituting \( g = 9.81 \, \text{m/s}^2 \) and using \( t = \frac{5 + \sqrt{10}}{3} \) (taking the positive root):
\[ H = \frac{1}{2} \cdot 9.81 \cdot \left(\frac{5 + \sqrt{10}}{3}\right)^2 \]
- Calculating this gives \( H \approx 36.27 \, \text{m} \).

### Final Answer:
The height of the tower \( H \) is approximately \( 36.27 \, \text{m} \).