If a stone dropped from the top of a tower travels half, of the height of the tower during last second of its fall, the time of fall is (in seconds)

Options

A. `3 + sqrt(2)`
B. `4 + sqrt(2)`
C. `2 -sqrt(2)`
D. `2 + sqrt(2)`

Answer

To solve the problem, we need to find the time of fall of a stone dropped from the top of a tower, given that it travels half of the height of the tower during the last second of its fall.

### Step-by-Step Solution:

1. **Define Variables**:
Let the height of the tower be \( h \) and the time of fall be \( n \) seconds.

2. **Use the Kinematic Equation**:
The distance fallen in \( n \) seconds can be expressed using the kinematic equation:
\[ h = ut + \frac{1}{2}gt^2 \]
Here, the initial velocity \( u = 0 \) (since the stone is dropped), and \( g \) (acceleration due to gravity) is taken as \( 10 \, \text{m/s}^2 \). Thus, the equation simplifies to:
\[ h = \frac{1}{2} \times 10 \times n^2 = 5n^2 \]

3. **Distance Fallen in the Last Second**:
The distance traveled during the last second of the fall can be calculated using the formula for the distance traveled in the \( n \)-th second:
\[ S_n = u + \frac{g}{2}(2n - 1) \]
Substituting \( u = 0 \) and \( g = 10 \):
\[ S_n = 0 + \frac{10}{2}(2n - 1) = 5(2n - 1) \]
According to the problem, this distance is equal to half of the height of the tower:
\[ S_n = \frac{h}{2} \]
Therefore, we can write:
\[ 5(2n - 1) = \frac{h}{2} \]

4. **Substituting for \( h \)**:
We already found that \( h = 5n^2 \). Substituting this into the equation gives:
\[ 5(2n - 1) = \frac{5n^2}{2} \]
Simplifying this, we can cancel \( 5 \) from both sides:
\[ 2(2n - 1) = \frac{n^2}{2} \]
Multiplying through by \( 2 \) to eliminate the fraction:
\[ 4(2n - 1) = n^2 \]
Expanding this:
\[ 8n - 4 = n^2 \]
Rearranging gives:
\[ n^2 - 8n + 4 = 0 \]

5. **Solving the Quadratic Equation**:
We can solve this quadratic equation using the quadratic formula:
\[ n = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
Here, \( a = 1, b = -8, c = 4 \):
\[ n = \frac{8 \pm \sqrt{(-8)^2 - 4 \cdot 1 \cdot 4}}{2 \cdot 1} = \frac{8 \pm \sqrt{64 - 16}}{2} = \frac{8 \pm \sqrt{48}}{2} \]
Simplifying \( \sqrt{48} = 4\sqrt{3} \):
\[ n = \frac{8 \pm 4\sqrt{3}}{2} = 4 \pm 2\sqrt{3} \]

6. **Selecting the Valid Solution**:
Since time cannot be negative, we take:
\[ n = 4 + 2\sqrt{3} \]
This is the time of fall in seconds.

### Final Answer:
The time of fall is \( n = 4 + 2\sqrt{3} \) seconds.